av Matthew Lichtsinn 1 år siden
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We invert and multiply fractions instead of just dividing them because it's easier to find answer when we change one of the fractions into its reciprocal and multiply rather then dividing them straightforward.
If we were to just divide fractions, we would end up solving it like, for example: 6/8 ___ 2/4 which can get pretty confusing. We would end up getting the same answer if we solved it this way, but multiplying those fractions with the second fraction being its reciprocal is the best way to solve this equation.
To solve this: 6/8 ___ 2/4, we would take the 8 from 6/8 and multiply with the fraction 8/8 or 1. The denominator of 6/8, and the numerator of 8/8 cancels out, leaving us to multiply the numerator of 2/4 with the denominator of 8/8. This will give us 6/16/4. Next, we would take the 4 and multiply 6/16/4 by 4/4. The two denominators of 4 cancel each other out, so, when we multiply the numerators 6 and 4, we end up with the fraction 24/16.
To solve this:
6/8
___
2/4,
we would take the 8
from 6/8 and multiply
with the fraction 8/8 or 1.
The denominator of 6/8, and the numerator of
8/8 cancels out, leaving us to multiply the numerator
of 2/4 with the denominator of 8/8. This will give us
6/16/4. Next, we would take the 4 and multiply 6/16/4
by 4/4. The two denominators of 4 cancel each other out,
so, when we multiply the numerators 6 and 4, we end up
with the fraction 24/16.
Ex. 6/8 divided by 2/4 Abstractly, we get: 6/8 x 4/2 = 24/16 = 6/4 = 3/2 = 1 1/2 Having a rectangle with a 6/8 fraction and another rectangle with the 2/4 fraction, we add 4 columns to the 6/8 rectangle and 8 rows in the 2/4 rectangle, leaving us with two rows and two columns unshaded. We now have 24 shaded parts in the 6/8 rectangle, and 16 shaded parts in the 2/4 rectangle.
Ex. 6/8 divided by 2/4
Abstractly, we get:
6/8 x 4/2 = 24/16 = 6/4 = 3/2 = 1 1/2
Having a rectangle with a 6/8 fraction and
another rectangle with the 2/4 fraction, we
add 4 columns to the 6/8 rectangle and 8 rows
in the 2/4 rectangle, leaving us with two rows and
two columns unshaded. We now have 24 shaded parts
in the 6/8 rectangle, and 16 shaded parts in the 2/4 rectangle.
Ex. 6/8 divided by 2/4 (continued) Crossing out 16 parts in the 6/8 rectangle, we have 1 group of 16 parts which is 1. There are now 8 remaining parts, which then, leaves us with the fraction of 8/16 or 1/2. Bringing the 1 group of 16 and the 8/16 remaining parts together, we get a mixed number of 1 1/2
Ex. 5/6 divided by 2/3 First, solving this abstractly, we get 5/6 x 3/2 = 15/12 = 5/4 = 1 1/4. Pictorially, we draw out two big rectangles on either side of each other drawing the 5/6 fraction in the first rectangle in rows. In the second rectangle, we draw out the 2/3 fraction in columns. Next, we add 3 columns in the 5/6 rectangle and 6 rows in the 2/3 fraction. We now have 15 parts shaded in the 5/6 rectangle with one row left unshaded, and 12 parts shaded in the 2/3 rectangle with one column left unshaded.
Ex. 5/6 divided by 2/3
First, solving this abstractly,
we get 5/6 x 3/2 = 15/12 = 5/4 = 1 1/4.
Pictorially, we draw out two big rectangles
on either side of each other drawing the 5/6
fraction in the first rectangle in rows. In the second
rectangle, we draw out the 2/3 fraction in columns.
Next, we add 3 columns in the 5/6 rectangle and 6 rows
in the 2/3 fraction. We now have 15 parts shaded in the 5/6 rectangle with one row left unshaded, and 12 parts shaded in the 2/3 rectangle with one column left unshaded.
Ex. 5/6 divided by 2/3 (continued) Then, we would cross out the number of parts in the 5/6 rectangle so that it matches the number of parts in the 2/3 rectangle. In this case, we would cross out 12 parts out of 15 from the 5/6 rectangle which gives us 1 group of 12 parts which equals 1. Now, we have 3 remaining parts shaded, so, we would have the fraction 3/12. Simplified, that's 1/4. Bringing the 1 group of 12 to the 3/12 parts remaining, we now have a mixed number of 1 1/4.
Ex. 5/6 divided by 2/3 (continued)
Then, we would cross out the number
of parts in the 5/6 rectangle so that it matches
the number of parts in the 2/3 rectangle. In this
case, we would cross out 12 parts out of 15 from the
5/6 rectangle which gives us 1 group of 12 parts which
equals 1. Now, we have 3 remaining parts shaded, so, we
would have the fraction 3/12. Simplified, that's 1/4. Bringing
the 1 group of 12 to the 3/12 parts remaining, we now have
a mixed number of 1 1/4.
Pictorially: This can be shown in a similar way from when we multiply fractions.
Ex. 1/3 dividing by 1/2 To figure this out pictorially, we first have to ask ourselves, in this case, how many 1/2s fit in or cover 1/3s? If one hexagon made a whole: The rhombus makes 1/3, and the trapezoid equals 1/2. So, we need to find out how much of the trapezoid covers the rhombus. As we first draw the rhombus, then draw a triangle at the end to make a trapezoid, we see that the trapezoid covers 2/3 of the rhombus. So, 1/3 x 2/1 = 2/3
Ex. 1/3 dividing by 1/2
To figure this out pictorially,
we first have to ask ourselves,
in this case, how many 1/2s fit in
or cover 1/3s?
If one hexagon made a whole:
The rhombus makes 1/3, and the
trapezoid equals 1/2. So, we need to find
out how much of the trapezoid covers the
rhombus. As we first draw the rhombus,
then draw a triangle at the end to make a
trapezoid, we see that the trapezoid covers
2/3 of the rhombus. So, 1/3 x 2/1 = 2/3
Ex. 1 1/2 divided by 1/6 How many 1/6 fit in or cover 1 1/2? If two hexagons equal a whole: The rhombus equals 1/6, and since two hexagons won't make it to 1 1/2, we add another hexagon and draw in three rhombi in each hexagon. With 3 rhombi in each hexagon, we end up with a total of 9 rhombi. So, 1 1/2 or 3/2 x 6/1 = 9
Ex. 1 1/2 divided by 1/6
How many 1/6 fit in or cover
1 1/2?
If two hexagons equal a whole:
The rhombus equals 1/6, and since
two hexagons won't make it to 1 1/2,
we add another hexagon and draw in
three rhombi in each hexagon. With 3
rhombi in each hexagon, we end up with
a total of 9 rhombi. So, 1 1/2 or 3/2 x 6/1 = 9
Dividing Fractions with Pattern Blocks Video:
Abstractly: We do things a little differently when divide fractions abstractly.
Ex. 1/3 divided by 1/2 To solve this, we would turn the division sign into a multiplication sign and put in the reciprocal of 1/2 which is 2/1. Multiplying 1/3 x 2/1, we get 2/3
Ex. 1 1/2 divided by 1/6 First changing the mixed number into an improper fraction, we get 3/2 x 6/1 = 18/2 which equals 9.
Pictorially: When using the fraction bar model pictorially, we would draw out one box, divide it into a certain number of parts, and shade some parts in depending on the fractions given.
Ex. 3/4 x 1/2 Drawing one box, we would first take the 1/2 fraction, divide the box into two parts horizontally, then shade in 1 of them. Next, we take the 4 from 3/4, then divide it into four columns. We now have a total of 8 parts in the box with 4 shaded, but since we only need 3, in order to solve the problem correctly, we cross out one part. This will leave us with 8 parts with only 3 shaded. Therefore, 3/4 x 1/2 = 3/8
Ex. 3/4 x 1/2
Drawing one box,
we would first take the 1/2 fraction,
divide the box into two parts horizontally, then
shade in 1 of them. Next, we take the 4 from 3/4,
then divide it into four columns. We now have a total
of 8 parts in the box with 4 shaded, but since we only
need 3, in order to solve the problem correctly, we
cross out one part. This will leave us with 8 parts with
only 3 shaded. Therefore, 3/4 x 1/2 = 3/8
Ex. 3/8 x 1/3 Drawing one box, we would first take the 3 from 1/3 and divide the box into 3 rows shading in 1 row. Taking the 8 from 3/8, we would draw out 8 columns. There are now 24 total parts with 8 parts shaded. Since we only need 3 parts in order for the answer to make sense, we would cross out 5 of those shaded parts, leaving us with the fraction 3/24. Simplified, gives us 1/8. Therefore, 3/8 x 1/3 = 1/8
Ex. 3/8 x 1/3
Drawing one box, we would
first take the 3 from 1/3 and
divide the box into 3 rows
shading in 1 row. Taking the
8 from 3/8, we would draw out 8
columns. There are now 24 total parts
with 8 parts shaded. Since we only need 3 parts
in order for the answer to make sense, we would
cross out 5 of those shaded parts, leaving us with
the fraction 3/24. Simplified, gives us 1/8. Therefore,
3/8 x 1/3 = 1/8
Abstractly: This is used the same way as when we add and subtract fractions, and multiplying fractions with pattern blocks.
Ex. 3/4 x 1/2 = 3/8 This is the simplified fraction Ex. 3/8 x 1/3 = 3/24. Simplified, that's 1/8
Pictorially: When multiplying fractions pictorially, since we are using the pattern block model we would draw out certain pattern blocks depending on the fractions given.
Ex. 2/3 x 1/4 If two hexagons made a whole: The pattern block that represents 1/4 in this case is the trapezoid. So, we would draw one trapezoid in one of the hexagons. Now, we have to find out what 2/3rds of 1/4th is. In this case, 2 triangles fill in 2/3rds of the trapezoid, and since the triangle represents 1/12 when two hexagons are a whole, then the answer is 2/12. Simplified, we get 2/3 x 1/4 = 1/6
Ex. 2/3 x 1/4
If two hexagons made a whole:
The pattern block that represents
1/4 in this case is the trapezoid. So,
we would draw one trapezoid in
one of the hexagons. Now, we have to
find out what 2/3rds of 1/4th is. In this case,
2 triangles fill in 2/3rds of the trapezoid, and
since the triangle represents 1/12 when two
hexagons are a whole, then the answer is 2/12.
Simplified, we get 2/3 x 1/4 = 1/6
Ex. 4/12 x 1/2 First, I always find it easier to solve the problem abstractly before I draw it out. So, 4/12 x 1/2 = 4/24. Simplified, that's 1/6.
Ex. 4/12 x 1/2 (continued) If four hexagons made a whole: The pattern block that represents 1/12 is the rhombus, and since the fraction in the equation calls for 4/12, then we would draw in 4 rhombi in the 4 hexagons. Now, we have to figure out what's half of 4/12. To do so, we'll draw in 1 triangle in each of the rhombi which would give us 4 triangles. Finally, because the fraction that resembles triangles, if four hexagons make a whole, is 1/24, then the total fraction would be 4/24. Simplified, that's 1/6. So, 4/12 x 1/2 = 1/6
Ex. 4/12 x 1/2 (continued)
If four hexagons made a whole:
The pattern block that represents
1/12 is the rhombus, and since the
fraction in the equation calls for 4/12,
then we would draw in 4 rhombi in the
4 hexagons. Now, we have to figure out
what's half of 4/12. To do so, we'll draw in
1 triangle in each of the rhombi which would
give us 4 triangles. Finally, because the fraction
that resembles triangles, if four hexagons make a
whole, is 1/24, then the total fraction would be 4/24.
Simplified, that's 1/6. So, 4/12 x 1/2 = 1/6
Multiplying Fractions with Pattern Blocks Video:
Abstractly: When we multiply fractions abstractly, we just multiply straight across and simplify when needed.
Ex. 2/3 x 1/4 Multiplying straight across, we get 2/12. This fraction can be simplified by dividing both the denominator and numerator by 2 which would give us a total of 1/6
Ex. 3/5 x 2/3 Multiplying these two fractions, we get 6/15. This can't be simplified anymore, so this our final answer.
Ex. 3/8 and 2/5 We would draw the first square dividing it into 8 parts vertically, shading in 3 parts. Then have the second square divided into 5 pieces horizontally, shading in 2 parts.
How we would solve this is, we would multiply the two denominators on either side. To do this pictorially, we would take the denominator 5 from 2/5, and draw in 5 rows in the 3/8 box, which now shows 15 parts shaded with a total of 40 parts. Abstractly, we multiply 5 by 3/8 which gives us a new fraction of 15/40.
How we would solve this is, we would
multiply the two denominators on either side.
To do this, we would multiply 5 by 3/8 which gives us
a new fraction of 15/40. Next, we would multiply 8 by 2/5
giving us 16/40. So, we can see now that 2/5 is larger than 3/8.
Next, we would take the 8 from 3/8 and draw in 8 columns in the 2/5 box which would show 16 parts shaded with a total of 40 parts. Then, we would multiply 8 by 2/5 giving us 16/40. So, we can see now that 16/40 is larger than 15/40.
Comparing Fractions Video:
Ex. 4 rod In this case, two 2 rods are used in order to fit in perfectly with the 4 rod. That means that 2 is 1/2 of 4.
Ex. 9 rod Putting three 3 rods under the 9 makes them fit in perfectly. Since three 3 rods make up 9, 3 is 1/3 of 9.
Ex. 6 rod Three 2's or two 3's make up the 6 rod. So, 2 is 1/3 of 6, and 3 is 1/2 of 6.
Ex. 10 rod Two 5's fit into the 10 rod. So, 5 is 1/2 of 10.
Ex 5 rod Five 1 rods can only fit into the 5 rod. So, 1 is 1/5 of 5. If we were to switch that up, one 5 rod perfectly fits in with five 1's, So, 5 is 5/1 of 1.
Cuisenaire Rods Video
Ex. 4 / 11, 16 / 48, 34 / 86
Improper Fractions: Fractions where the numerator is larger than the denominator.
Ex. 14 / 6, 24 / 7, 30 / 10
Mixed Number/Mixed Fractions: A way to represent an improper fraction as a whole number with a proper fraction.
Ex. 24 / 7 To change this improper fraction into a mixed number, we would divide 24 by 7. 7 goes into 24 three times. 7 x3 = 21 24 -21 = 3. So, our mixed number, keeping the original denominator is 3 3/7
Ex. 6 3/4 This can be changed into an improper fraction by multiplying the whole number with the denominator. In this case, 6 x 4 = 24. Then, we add the total with the numerator. 24 + 3 = 27. So, our improper fraction, keeping 4 as the denominator is 27 / 4.
Ex. 0.2307692308 = (3 / 13) Ex. 0.5714285714 = (4 / 7) Ex. 0.2777777778 = (5 / 18)
Ex. 0.625 = (5 / 8) Ex. 0.1875 = (3 / 16) Ex. 0.21875 = (7 / 32)
Ex. There are 12 girls and 20 boys in Mr. Green's class. If 25% of the girls and 30% of the boys are going on a trip to the history museum, what percentage of the entire class is going on the trip? To solve this, we would multiply the percentage of boys and girls by the number of boys and girls. So, we would solve this as follows: 25% x 12 = 3 and 30% x 20 = 6. Next, we would add up the two total numbers and the total number of girls and boys. 3 + 6 = 9 and 20 + 12 = 32. We would then divide 9 and 32 giving us 0.28125. Rounding to the nearest whole number, we get 28%. So, 28% of the whole class is going on the trip.
Ex. There are 12 girls and 20 boys in Mr. Green's class.
If 25% of the girls and 30% of the boys are going on a
trip to the history museum, what percentage of the entire
class is going on the trip?
To solve this, we would multiply the percentage of boys and girls by the number of boys and girls. So, we would solve this as follows: 25% x 12 = 3 and 30% x 20 = 6. Next, we would add up the two total numbers and the total number of girls and boys. 3 + 6 = 9 and 20 + 12 = 32. We would then divide 9 and 32 giving us 0.28125. Rounding to the nearest whole number, we get 28%. So, 28% of the whole class is going on the trip.
Ex. Jason bought a truck 1 year ago, but now, it has decreased by $16,243 which is 25% of the price he paid for it. How much did Jason pay for the truck to the nearest dollar? Because the original price decreased by $16,243, then the original price will be greater than $16,243. Since there is a 25% decrease, we would write the equation as follows. Let m equal the original price: $16,243 = 0.25 x m Next, we would divide the current price by the percentage. 16,243 / 0.25 = 0.25 x m / 0.25 = 64,972. Therefore, $64,972 is the original price Jason paid for his truck.
Ex. Jason bought a truck 1 year ago, but now,
it has decreased by $16,243 which is 25% of the
price he paid for it. How much did Jason pay for the
truck to the nearest dollar?
Because the original price decreased by $16,243, then
the original price will be greater than $16,243. Since there
is a 25% decrease, we would write the equation as follows.
Let m equal the original price:
$16,243 = 0.25 x m
Next, we would divide the current price by the percentage.
16,243 / 0.25 = 0.25 x m / 0.25 = 64,972. Therefore, $64,972
is the original price Jason paid for his truck.
Ex. A vase usually cost $95, but today, it's on sale for a 20% discount. What is the sale price of the vase? $95.00 x 20% = $19.00. So, the sale price of the vase is $19.00
Ex. Eddie wanted to jog for 25 minutes but he decreased it to 15 minutes. What was the percent of the decrease? We would first subtract 25 by 15 which is 10, then divide the original time, 25 minutes by 10 which gives us a result of 2.5%
Ex. ___ 0.28 Since there are 2 digits after the repeating decimal, we would multiply the repeating decimal by 10^2. Having the letter n equal the repeating decimal, we would have the following equation: ___ 100n = 28.28 ___ - n = 0.28 = 99n = 28 We now have the result of 99 / 28 which gives a repeating decimal of 0.28
Ex. ___
0.28
Since there are 2 digits after the repeating decimal,
we would multiply 10^2. Having the letter n equal the
repeating decimal, we would have the following equation:
___
100n = 28.28
___
- n = 0.28 =
99n = 28
We now have the result of 99 / 28 which gives a repeating
decimal of 0.28
Ex. ___ 1.327 To figure this equation out, we would turn this into a problem we already know how to solve, having the repeated decimal right after the decimal point. To solve that, we would multiply the original repeated decimal by 10, which would give us: ___ 13.27 Now, since there are 2 digits in the repetend, we would multiply the repeating decimal by 10^2 or 100 multiplied by 10. This equals ___ 1327.27
Ex. ___
1.327
To figure this equation out,
we would turn this into a problem
we already know how to solve, having
the repeated decimal right after the decimal point.
To solve that, we would multiply the original repeated
decimal by 10, which would give us:
___
13.27
Now, since there are 2 digits in the repetend, we would multiply the repeating decimal by 10^2 or 100 multiplied by 10. This equals
___
1327.27
Now we can solve the following equation: ___ 1000n = 1327. 27 ___ - 10n = 13.27 990n = 1314 = 1314 / 990, or 73 / 55
Ex. 7/12 = 0.5833333333 The 3 in the thousandth place starts the repetition. So it would be shown as, __ .583
Ex. 0.4444444444 = (4 / 9) Ex. 0.636363636 = (7 / 11) Ex. 0.409090909 = (9 / 22) *Notice in the last example, the pattern begins in the hundredths place instead of in the tenths place.
Fun Fact: All members of the "1/11" family show a repeated decimal with multiples of 9. 1/11 = 0.0909090909 2/11 = 0.1818181818 3/11 = 0.2727272727 4/11 = 0.3636363636 5/11 = 0.4545454545 6/11 = 0.5454545454 7/11 = 0.6363636363 8/11 = 0.7272727272 9/11 = 0.8181818181 10/11 = 0.9090909090
All members of the "1/9" family show a repeated decimal of the same number. 1/9 = 0.1111111111 2/9 = 0.2222222222 3/9 = 0.3333333333 4/9 = 0.4444444444 5/9 = 0.5555555555 6/9 = 0.6666666666 7/9 = 0.7777777777 8/9 = 0.8888888888
All members of the "1/7" family have the same digits after the decimal in different combinations. 1/7 = 0.1428571429 2/7 = 0.2857142857 3/7 = 0.4285714286 4/7 = 0.5714285714 5/7 = 0.7142857143 6/7 = 0.8571428571
To determine where the pattern begins, there will be a bar over certain numbers telling us that those digits repeat after the decimal. Ex. ____ .209 In this example, the .209 is repeated. So, it would go, 0.209209209209 Ex. ___ .209 The digits 0 and 9 are repeated. So, it would go, 0.2090909090909 Ex. __ .209 Only the 9 is repeated. So, it would go, 0.2099999999999
To determine where the pattern begins,
there will be a line over certain numbers telling us
there are more digits after the decimal.
Ex. ____
.209
In this example, the .209 is repeated. So, it would go,
0.209209209209
Ex. ___
.209
The digits 0 and 9 are repeated. So, it would go,
0.2090909090909
Ex. __
.209
Only the 9 is repeated. So, it would go,
0.2099999999999
Repeating Decimals Video
Ex. 0.3 x 0.8 For this equation, we would use a 10x10 grid, count 3 units in the first column and 8 units in the top row, and then fill out the area. Counting all of the shaded units would give us a result of 0.24 or 24/100
Decimals and Grids Video
Ex. 0.41 Since there are 2 digits after the decimal, we would have a 10x10 grid displaying 100 unit cubes. We would simply fill in 41 unit cubes since the decimal 0.41 is also known as 41 /100
Associative Property: Rearranging the parenthesis in an equation gives you the same result Ex. (a + b) + c, a + (b + c)
Multiplicative Identity Property: When you multiply a certain number by 1, you would get a result of that certain number. Ex. 5 x 1 = 5, 10 x 1 = 10
Distributive Property: Multiplying the outside number with each number in the parenthesis. Ex. 5 (4 + 7) = 20 + 35 = 55.
Inverse Property of Multiplication: Every real number besides 0 has a reciprocal The product of any number and its reciprocal is one. Ex. 4 x 1/4 = 1, 8 x 1/8 = 1
1/5 = 0.2 or .2 2/5 = 0.4 or .4 3/5 = 0.6 or .6 4/5 = 0.8 or .8
2/9 = .2 continued 5/9 = .5 continued 7/9 = .7 continued 8/9 = .8 continued
2/11 = .18 continued 3/11 = .27 continued 4/11 = .36 continued 6/11 = .54 continued
1 / 5 = 0.2 2 / 8 = 0.25 3 / 4 = 0.75 4 / 8 = 0.5 Based on the decimals, the order of these fractions from smallest to largest are: 1/5, 2/8, 4/8, 3/4
If we were to order these fractions from largest to smallest, it would go: 3/4, 4/8, 2/8, 1/5
Smallest to Largest: 1 / 3 = 0.3 continued 4 / 6 = 0.6 continued 2 / 7 = 0.29 continued 5 / 9 = 0.5 continued The order is: 2/7, 1/3, 5/9, 4/6 Largest to Smallest: 4/6, 5/9, 1/3, 2/7
Natural Numbers: Counting numbers (1,2,3,4,5...) Ex. 60
Rational Numbers: Any number that can be written as a fraction of two integers. Ex. -0.4, 60, -3/4, 5.6, 3 1/3
Irrational Numbers: A number that can't be written as a fraction. Ex. pi, or e
Imaginary Numbers: Ex. The square root of -1
Integers: Whole numbers positive or negative Ex. 60, -60,
Ex. -6 x 3 Since we can't have a negative number of groups, we would take the positive number which is 3 and have 3 groups of 6 negative charges/chips. Adding all of the 6 charges up in each group gives us 18 negatives. So, -6 x 3 = -18
Ex. 8 x 5 Since both numbers are positive, we can either have 5 groups of 8 positives, or 8 groups of 5 positives. I would personally prefer to have a smaller number of groups with a bigger number of positives or negatives. Adding up all of the positives in each group, we would get 40 positives. So, 8 x 5 = 40
Charged Field Model Addition & Subtraction Video
Ex. -7 + 4 In this example, we would draw out 7 negative charges or chips and 4 positive charges or chips. To find the result we would cross out 4 groups where the negatives and positives meet. When negative or positive charges/chips, in this case, negative charges/chips are alone, we would count how many charges or chips are left. In this example, there are only 3 negative charges/chips left. So, -7 + 4 = -3
Ex. 5 + -6 In this example, we would have 5 positives and 6 negatives lined up in two rows. Next, we would cross out 5 groups, where the negative and positive charges/chips meet. We are now left with 1 negative charge/chip. So, 5 + -6 is -1
Ex. 3 - 8 To solve this, we would have 3 positives and 8 negatives. We would then have to cross out groups of these charges/chips. This will leave us with 5 negative charges/chips. So, 3 - 8 = -5 1 - (-2) In this equation, we would have 1 positive and two negatives. Then we put 2 positives over the two negatives, and since we are taking away -2, we would cross out the two negatives leaving us with 3 positives. So, 1 - (-2) = 3
Ex. -2 - 4 In this example, we would draw out 2 negative charges/chips and since we don't have -4 charges/chips, we would have to draw more charges/chips so that we would end up with 4 negative charges/chips. In this case, we would only draw two more negatives to get -4. Now what we have to do is draw two positives below the two negatives we have added. But since we can't take away 4 negatives with 2 positives, we would have to add two more positives to make four positives. Then, we would have to add two more negatives so that the negatives and positives line up. Next, we have to cross out all the positives and add up all of the negatives. This will give us 6 negatives. So, -2 - 4 = -6
Dividing Real Numbers: When we divide, we are trying to find what is called the quotient. 12 / 4 = 3 16 / -2 = -8 -60 / 5 = -12 -45 / -9 = 5
Multiplying Real Numbers: When we multiply, we are trying to find what is called the product. 4 x 3 = 12 8 x -2 = -16 -12 x 5 = -60 -9 x -5 = 45
Ex. 10 - 13 Starting with the 10 on the number line, we would go to the left 13 times landing on -3. So, 10 - 13 = -3 Ex. -1 - 4 Starting at -1, we go to the left 4 times landing on -5 So, -1 - 4 = -5
Ex. 7 - 9 Starting at 7, we go to the left 9 times landing on -2. So, 7 - 9 = -2 Ex. 15 - 8 Starting at 15, we go to the left 8 times landing on 7. So, 15 - 8 is 7.
Ex. 5 + -2 In this example, we would start at 5 and since the 2 is negative, we would go to the left 2 times on the number line landing on 3. So, 5 + -2 = 3 Ex. -4 - (-3) In this example, when we see two negative signs on either side of the parenthesis, the number inside the parenthesis becomes positive. So the equation would be -4 + 3. Then we would move to the right 3 times giving us -1.
Ex. -6 + -3 We would start at -6, and since the 3 is negative, we would keep going to the left landing on -9. So, -6 + -3 = -9 Ex. -2 + 1 - 7 To solve this equation, we would start at -2 and go to the right 1 landing on -1 So, we are left with the equation -1 - 7. In this case, starting at -1 we would go back to the left 7 times landing on -8. So, -2 + 1 - 7 = -8
Colored Rods: Ex. (3,4)
To find the least common multiple
between these numbers, we would take
a 3 cm. rod and a 4 cm. rod, collect a certain number of them,
and build trains of these two rods so that they evenly line up with each other. In this case, we would have three 4 cm. rods and four 3 cm. rods evenly lined up together. Next, we would take longer/shorter rods and find out what rods evenly line up with the three 4 cm. and four 3 cm. rods. In this example, a 10 cm. rod and a 2 cm. rod perfectly fit with the three 4 cm and four 3 cm. rods. Adding the 10 cm. and 2 cm. rods together gives us a length of 12. Therefore, 12 is the least common multiple between 3 and 4.
LCM: Colored Rods Method Video
Prime Factorization: Ex. (60,24) We would do the same thing as we did when finding the greatest common factor. But this time, we would look for the largest exponents and numbers in the two equations to find the least common multiple. In this example, 2^3 is the largest exponent, and then we just include the 3 and 5 and multiply. We end up with a total of 120. Therefore, the least common multiple between 60 and 24 is 120.
Prime Factorization: Ex. (60,24)
We would do the same thing as we
did when finding the greatest common factor.
But this time, we would look for the largest exponents
and numbers in the two equations to find the least common
multiple. In this example, 2^3 is the largest exponent, and then we just include the 3 and 5 and multiply. We end up with a total of 120. Therefore, the least common multiple between 60 and 24 is 120.
The Intersection of Sets: Ex. (32, 44) To find the least common multiple between these two numbers, we would first have to find the multiples of each number and then find the least number that goes into 32 and 44. M32 {32,64,96,128,160,192,224,256,288,320,352,384...} M44 {44,88,132,176,220,264,308, 352,396...} In this case, 352 is the first multiple the two numbers have in common. Therefore, between 32 and 44, 352 is the least common multiple.
The Intersection of Sets: (32, 44)
To find the least common multiple
between these two numbers, we would first
have to find the multiples of each number and then
find the least number that goes into 32 and 44.
M32 {32,64,96,128,160,192,224,256,288,320,352,384...}
M44 {44,88,132,176,220,264,308, 352,396...}
In this case, 352 is the first multiple the two numbers have
in common. Therefore, between 32 and 44, 352 is the least
common multiple.
Number Line: Ex. (3,4) This method is mainly used when finding the least common multiple between smaller numbers. How we would use this method we would start at 0, pick any of the two numbers given to figure out first, and then go to right counting by whatever the number you chose to solve for first. In this case, we can take the 4 and go to the right of the number line skipping over 3 numbers, and counting by 4 landing on a certain number. Then, we can take the 3 and do the same thing but this time, skip over 2 numbers and count by 3.
Number Line: (3,4)
This method is mainly used when
finding the least common multiple between
smaller numbers. How we would use this method
we would start at 0, pick any of the two numbers given
to figure out first, and then go to right counting by whatever
the number you chose to solve for first. In this case, we can take the 4 and go to the right of the number line skipping over 3 numbers, and counting by 4 landing on a certain number. Then, we can take the 3 and do the same thing but this time, skip over 2 numbers and count by 3.
Once we have done that, we would then find out where on the number line both numbers similarly landed while in the process of going to the right on the number line. In this example, 12 is the number that both 3 and 4 landed on while going to the right. So, 12 is the least common multiple of 3 and 4.
Division-by-Primes: Ex. (75,120) To find the least common multiple between these two numbers, we would first take the smallest prime number: 2, and divide it by 75 and 120. Since 75 is not divisible by 2, we would just bring down the 75 and divide 120 by 2 which is 60. So, we now have 75 and 60 divided by 2. We would bring down the 75 again and divide 60 by 2 which is 30. Bringing down the 75 once more, we then divide 30 by 2 which is 15. Now, since neither 75 nor 15 can be divisible by 2, we would take the next prime number: 3, and divide these two numbers by 3. 75/3 = 25 and 15/3 = 5. We now have 25 and 5 divided by the next prime number: 5 since 3 can't divide into 25 or 5. 25/5 = 5 and 5/5 = 1. Taking the 5 and 1 and dividing it by 5 again, we end up with 1 and 1.
Division-by-Primes: Ex. (75,120)
To find the least common multiple between these two numbers, we would first take the smallest prime number: 2, and divide it by 75 and 120. Since 75 is not divisible by 2, we would just bring down the 75 and divide 120 by 2 which is 60. So, we now have 75 and 60 divided by 2. We would bring down the 75 again and divide 60 by 2 which is 30. Bringing down the 75 once more, we then divide 30 by 2 which is 15. Now, since neither 75 nor 15 can be divisible by 2, we would take the next prime number: 3, and divide these two numbers by 3. 75/3 = 25 and 15/3 = 5. We now have 25 and 5 divided by the next prime number: 5 since 3 can't divide into 25 or 5. 25/5 = 5 and 5/5 = 1. Taking the 5 and 1 and dividing it by 5 again, we end up with 1 and 1.
Once this is all figured out, we would then determine how many of each prime numbers it took to get to the bottom. In this case it took three 2's, one 3, and two 5's to get to the bottom of this equation. We would then multiply 2^3 x 3 x 5^2 which would give us 600. So, 600 is the least common multiple between 75 and 120.
Colored Rods: Ex. (6,8)
This method is more beneficial
when finding the greatest common factor
between single-digit numbers. To figure this out,
we would take a 6 cm. rod and an 8 cm. rod, then try to
find smaller rods that perfectly fit with the two rods.
In this case, three 2 cm. rods perfectly fit with the 6 cm. rod,
but since the 8 cm. rod is longer, the 3 rods can't be used to build the 8 cm. rod, so we would have to add another 2 cm. rod. This leaves us with four 2 cm. rods perfectly fitting with the 8 cm. rod. Other rods that are longer like 5 cm. rods
won't fit with the 6 or 8 cm. rod. A 6 cm. rod won't fit with the 8 cm. rod, so we would have 2 cm. rods fit with both rods. Therefore, 2 is the greatest common factor between 6 and 8.
Prime Factorization: Ex. (60,24) We have to find the factors of both of these numbers by creating a factor tree for each. 5 and 12 are factors that can be multiplied together to get 60. 2 and 6 are factors of 12, and 2 & 3 are factors that go into 6. So, we are left with the equation, 2^2 x 3 x 5. For 24, 3 and 8 are the two factors that go into 24. 2 and 4 are factors that go into 8, and 2 & 2 are the factors that go into 4. So, we are left with the equation 2^3 x 3.
Prime Factorization: Ex. (60,24)
We have to find the factors of both of
these numbers by creating a factor tree for each.
5 and 12 are factors that can be multiplied together
to get 60. 2 and 6 are factors of 12, and 2 & 3 are factors
that go into 6. So, we are left with the equation, 2^2 x 3 x 5.
For 24, 3 and 8 are the two factors that go into 24. 2 and 4 are
factors that go into 8, and 2 & 2 are the factors that go into 4.
So, we are left with the equation 2^3 x 3.
Once we figure out both equations, we would then take the smallest exponents and numbers combine them and then multiply. In this case 2^2 is the smallest exponent and since there is no exponents of 3 in either factor tree, we would just take the 3 and multiply it by 2^2 which equals 12. So, the greatest common factor between 60 and 24 is 12.
Once we figure out both equations,
we would then take the smallest exponents and numbers
combine them and then multiply. In this case 2^2 is the smallest exponent and since there is no exponents of 3 in either factor tree, we would just take the 3 and multiply it by 2^2 which equals 12. So, the greatest common factor between 60 and 24 is 12.
Intersection of Sets: (132, 244) This method may take a while, since we are trying to find the many numbers that divide into these two numbers. My suggestion would be to find at least 10 numbers that divide into each of these two numbers. D132 {1,2,3,4,6,11,12,22,33,44,66,132...} D244 {1,2,4,61,244...}
Once we find the majority of the numbers that can divide into each number, we would compare each set and find out what numbers they have in common. In this example, 132 and 244 can both be divisible by 1, 2, and 4. Since we are trying to find the greatest common factor, we would look for the largest number that can divide into 132 and 244. In this case, 4 is the largest number that can divide into both. So, 4 is the greatest common factor
Once we find the majority of the numbers that can divide
into each number, we would compare each set and find out
what numbers they have in common. In this example, 132 and 244 can both be divisible by 1, 2, and 4. Since we are trying to find the greatest common factor, we would look for the largest number that can divide into 132 and 244. In this case, 4 is the largest number that can divide into both. So, 4 is the greatest common factor
For the divisor 9, the digits in a number must add up to 9 in order to be divisible by 9 Ex. 45: 4+5 = 9, 45/9 = 5, 63: 6+3 = 9, 63/9 = 7, 117: 1+1+7 = 9, 117/9 = 13
For the divisor 4, the last two digits of a number must be even in order to be divisible by 4. Ex. 120/4 = 30, 188/4 = 47, 204/4 = 51
Adding Methods in Different Base Systems: When adding with different base systems, for example, 33base 6 + 21base 6, you would add two numbers together, for instance, 3 + 1 is 4, and since 4 is a number in base 6, you would write 4. 3+2 is 5 so it would be 54.
When using the lattice algorithm, we would draw boxes below the numbers being added and draw a slash in each to divide the boxes in two to show the ten's and one's place of a number. For example, 64 + 52. We would draw two boxes underneath since there are 2-digit numbers. 4 + 2 is 6 so we would put the 0 on the top part of the box in the one's place and the 6 in the bottom corner. Then add 6 + 5 which is 11, so the 1 will be in the top part of the box in the ten's place and another 1 on the bottom. We would then add diagonally. 0 + 1 is 1 and then bring down the other 1 and 6 which equals 116.
Column addition is similar to standard algorithm. We would take two numbers, for instance 441 + 329, and put vertical lines separating each number putting it in the one's, ten's, and hundred's column. We would then simply add. 1 + 9 is 10, 4 + 2 is 6, and 4 + 3 is 7. Next, we take any 2-digit numbers and carry the ten's place number to the next column. So, we would carry the 1 in 10 and add 1 to 4 + 2 which is 6 + 1 is 7. So 441 + 329 = 770
Column addition is similar to the standard algorithm. We would take two numbers, for instance, 441 + 329, and put vertical lines separating each number putting it in the one's, ten's, and hundred's column. We would then simply add. 1 + 9 is 10, 4 + 2 is 6, and 4 + 3 is 7. Next, we take any 2-digit numbers and carry the ten's place number to the next column. So, we would carry the 1 in 10 and add 1 to 4 + 2 which is 6 + 1 is 7. So 441 + 329 = 770
When we use the opposite change method, we would add and subtract a certain number to any of the two numbers in the equation so that one of them ends in 0. For example, 398 + 277. First, we would determine which number is closer to a number that ends in 0. In this case, 398 would be the best choice, so we would add 2 to 398 to make 400, but we have to do the opposite of 277, so we would subtract 2 from 277 to make 275. 400 + 275 which is 675
When we use the opposite change method, we would
add and subtract a certain number to any of the two numbers in the equation so that one of them ends in 0. For example, 398 + 277. First, we would determine which number is closer to a number that ends in 0. In this case, 277 would be the best choice, so we would add 3 to 277 to make 280, but we have to do the opposite of 398, so we would subtract 3 from 398 to make 395. Next, we'll need to turn the 8 in 280 to a 0, so if we add 20 to 280, that would give us 300. Then, subtract 20 from 395 which is 375. So we get 375 + 300 which is 675.
A partial sum example would be, 365 + 244. You would first take the hundreds place value and write 300 then take the hundreds place value from the other number and write 200, for the rest, 60, 40, 5, and 4, and then add them all up which would give a total of 609.
Standard Algorithms are when you basically add the numbers in the same column. Like, 51 + 24 = 75.
Ex. 9 would be one line and 4 circles Ex. 16 would be two lines horizontally parallel, one line on the right side of the bottom line, and a circle on the side Ex. 23 is one big circle on top and three little circles below it Ex. 106 is one big line on top, one little line on the bottom, and a circle on the side.
Ex. 9 would be one line and 4 circles
Ex. 16 would be two lines horizontally parallel, one line on the right side of the bottom line, and a circle on the side
Ex. 23 is one big circle on top and three little circles below it
Ex. 106 is one big line on top, one little line on the bottom, and a circle on the side.
Ex. Three 1/4 fraction bars = 3/4 Ex. Two 1/7 fraction bars = 2/7 Ex. Eight 1/11 fraction bars = 8/11
When it comes to subtracting fractions using fraction bars, we would place one below the other and fill in the empty space below which will be our difference.
Ex. 6/9 - 1/3 Putting the 1/3 fraction bar below the 6/9 fraction bar, the fraction bar that easily fits in the empty space is 2/6 or 1/3. Therefore, 6/9 - 1/3 = 1/3
Subtracting Fractions with Fraction Strips Video:
Ex. 6/12 - 2/8 Putting the 2/8 fraction bar below the 6/12 fraction bar, we have to fill in the empty space with another fraction bar that fits in perfectly. In this case, the 1/4 fraction bar fits in evenly fits in the empty space. Therefore, 6/12 - 2/8 = 1/4
When it comes to adding fractions using fraction bars, we would place them side by side and figure out how much of one unit fraction bar is needed to evenly fit with the two fractions
Ex. 3/9 + 1/6 Lining these up, side by side, below, we would get 6/12. Simplified, we'll end up with 3/9 + 1/6 = 1/2
Adding Fractions with Fraction Strips Video:
Ex. 5/10 + 2/6 Lining up five 1/10s and two 1/6s fraction bars side by side, we place one fraction bar below. In this case, we would have ten 1/12s or 10/12. Since this fraction can be simplified, we would end up with 5/6. Therefore, 5/10 + 2/6 = 5/6
If one trapezoid made a whole: Hexagon = 2 Rhombus = 2/3 Triangle = 1/3
If two rhombi make a whole: Hexagon = 1 1/2 Trapezoid = 3/4 Triangle = 1/4
When using pattern blocks, a trapezoid equals 1/2, a rhombus equal 1/3, and a triangle equals 1/6, if 1 hexagon makes a whole.
If two hexagons made a whole: Hexagon = 1/2 Trapezoid = 1/4 Rhombus = 1/6 Triangle = 1/12
If three hexagons made a whole: Hexagon = 1/3 Trapezoid = 1/6 Rhombus = 1/9 Triangle = 1/18
If four hexagons made a whole: Hexagon = 1/4 Trapezoid =1/8 Rhombus = 1/12 Triangle = 1/24
Ex. 2/3 - 5/7 We first have one box with 3 rows with 2 of the rows shaded. Another box with 7 columns with 5 columns shaded. Adding 7 columns in the 2/3 box, we end up with 21 total parts with 14 parts shaded. Multiplying 7 by 2/3 gives us 14/21. Adding 3 rows to the 5/7 box gives us 21 parts total with 15 parts shaded. Multiplying 3 by 5/7 gives us 15/21. Subtracting the 14/21 by 15/21 gives us -1/21.
Ex. 4/5 + 1/4 We would draw one box drawing in 5 columns with 4 of those columns shaded, and another box drawing in 4 rows shading in 1 row. Next, we'll take the 4 from 1/4 and draw in 4 rows in the 4/5 box, which will give us 16 parts shaded with a total of 20 parts. Abstractly, we multiply 4 by 4/5 which also gives us 16/20. Then, we will take the 5 from 4/5 and draw in 5 columns in the 1/4 box, giving us 5 parts shaded with a total of 20. Multiplying 5 by 1/4 = 5/20.
Ex. 4/5 + 1/4
We would draw one box drawing
in 5 columns with 4 of those columns shaded,
and another box drawing in 4 rows shading in 1 row.
Next, we'll take the 4 from 1/4 and draw in 4 rows in the
4/5 box, which will give us 16 parts shaded with a total
of 20 parts. Abstractly, we multiply 4 by 4/5 which also gives us 16/20.
Then, we will take the 5 from 4/5 and draw in 5 columns in the
1/4 box, giving us 5 parts shaded with a total of 20. Multiplying 5 by 1/4 = 5/20.
Adding the two new fractions: 16/20 + 5/20, we get 21/20. Since this is an improper fraction, we can turn this into a mixed number. Dividing 21 by 20 gives us the mixed fraction of 1 1/20
Percentage Decrease: Discount - The amount by which an original amount priced has decreased Final price = original price - discount
Ex. Percentage Increase: Find the result when 40 is increased by 25%. To figure this out, we would multiply the percentage by the number given. In this case, we would write 0.25(40) = 10. Next, since this is an increase, we would add the new result to the original number. 40 + 10 = 50. Therefore, the result is 50.
Ex. Percentage Decrease Find the result when 35 is decreased by 20%. We would first do the same thing we did when finding the result of the increase. Multiplying the percentage with the number: 0.20(35) = 7. This time, we would subtract the new result from the original number. In this case, 35 - 7 = 28. Therefore, the result is 28.
Ex. From 30 to 18. 18 - 30 = -12 / 30 = -0.4 = -40% This is a percent decrease because 30 equals the original value, while 18 equals the new value, and since 18 is less than 30, the original value dropped by 40%.
Ex. From 38 to 55 55 - 38 = 17 / 38 = 45% This is a percent increase since the original value (38) is smaller than the new value (55). Therefore, the original value increased by 45%.
Percentage of Increase/Decrease Video
Ex. On Monday, 15 students signed up to go on a field trip. Then, on Tuesday, a total number of 28 students signed up for the field trip. What is the percentage change? 28 - 15 = 13/15 = 87%
Ex. In 2021, the cost of a wedding dress was $2,000. Now, the price of a wedding dress is $3,500. What is the percent change? 3500 - 2000 = 1500/2000 = 0.75 = 75%
Ex. Joanne had 20 pieces of candy, and she gave 11 pieces to her brother. What is the percent change? 20 - 11 = 9 9 (New Amount) - 20 (Original Amount) = -11/20 = -0.55 So, the percentage change is -55%
Ex. During a math test, one student completed 25 questions. Another student completed 10 questions less than the other student. What was the percentage change? 25 -10 = 15 15 - 25 = -10/ 25 = -0.4 which equals -40%.
Percent Change Video
Ex. 0.016 x 0.04 Simply multiplying, 0.016 x 0.04 = 0.00064. Since 0.016 has 3 digits after the decimal, and 0.04 has 2 digits after the decimal, we would have a result that has 5 digits after the decimal.
Ex. 3.45 + 1.22 Another way to solve this equation would be as follows: (3 + 4/10 + 5/100) + (1 + 2/10 + 2/100) = (3 + 1) + (4/10 + 2/10) + (5/100 + 2/100) = 4 + 6/10 + 7/100 = 4.67
Ex. 231,000,000
To figure out the scientific notation of this number,
we would have to place a decimal in between the first two
numbers that aren't 0. In this case, we would put the decimal
in between 2 and 3, so, the decimal we would get is 2.31.
Next, we would have to multiply it by 10 with either a positive or negative exponent. Since 2.31 is less than 231,000,000, then the exponent would be positive. To find what the exponent is, we would count how many places the decimal moved. In this case, the decimal moved to the left 8 times.
So, the equation would be 2.31 x 10^ (8)
Ex. .00045 To find the scientific notation of this number, the decimal would be placed between 4 and 5 giving us a decimal of 4.5. Now, since 4.5 is greater than .00045, then the exponent will be negative. Since the decimal moved 4 places to the right, the equation would be 4.5 x 10^ (-4)
Ex. 2.1 / 0.3
Using the base 10 blocks,
we would have 2 flats and 1 rod.
Next, we would break down the 2 flats
into 10 rods each. Counting the 1 rod, we
have a total of 21 rods. Now, for this equation,
instead of dividing these rods into 3 groups, we would
have to determine how many groups of rods there are in
order to get the answer. To do that, we would divide these rods into groups of 3. We then end up with a total of 7 groups. So, 2.1 / 0.3 = 7.
Dividing:
Ex. 1.4 / 7
With base 10 blocks,
we would have 1 flat and 4 longs.
Now what we are going to do is divide these
into 7 groups. To do that, we would substitute the
1 flat with 10 rods so we have a total of 14 rods.
We can now take these rods and divide them into 7 groups.
This would result in us getting 2 rods in each of the 7 groups.
So, 1.4 / 7 = 0.2
Dividing Decimals Using Base 10 Blocks Video
Ex. 2.5 x 1.3
Using base 10 blocks,
we would have 2 flats, and, this time,
5 rods beside the 2 flats since the 5 is in the tenths place
Then, we would have 3 rods under each flat. Now, since
we have an empty space in the far-right corner, we would
have to put unit cubes to fill it in. Since there are 3 rows of rods underneath the 2 flats and 5 columns of rows, we have 3 rows of 5 units of cubes. Using the table, we have a 2 in the 1's place, 11 in the 10's place, and a 15 in the 100's place. Taking
the 1 from 11 and adding to 2 gives us 3 in the 1's place. Then, taking the 1 from the 15 and adding it to 1 in the 10's place, we have 2 leaving us with 3.25. So, 2.5 x 1.3 = 3.25
Ex. 1.4 x 0.3 Using base 10 blocks, we would solve this equation differently. We would have 1 flat and 4 rods underneath. Then, looking at the 3 in 0.3, we would count up 3 columns in the flat, line it up with 4 rods, and draw a line to split each of them up leaving us with 3 rods and 12 unit cubes. Carrying the 1 from 12 and adding it to 3, we end up getting 0.42. So, 1.4 x 0.3 = 0.42
Multiplying Decimals Using Base 10 Blocks Video
Multiplying:
Ex. 4 x 1.3
To solve this using base-10 blocks,
since 4 is a whole number, we would have
4 flats or wholes in one row. Since 1.3 is a decimal, we would
first, look at the 1 and determine how many flats we need
to make a column. Since we have 1 flat from the 4 flats, we don't need to add a column. Next, we would take the 3, and because 3 is in the tenth place, we would have 3 rods under each of the 4 flats since 1 rod equals 1/10.
Then, we would draw a table separating it into 1's, 10s, and 100's place writing out how many flats we have, and how many rods we have. In this case, we have 4 in the 1's place resembling 4 wholes, and, counting up the number of rods, we end up with 12 in the 10's place. Because there are no unit cubes the number in the 100's place is 0. For the 10's place, since we can't have more than one-digit numbers in each place, we would take the 1 from 12 and add it to 4 leaving us with 5 in the one's place and 2 in the ten's place. Therefore, 4 x 1.3 = 5.2
Then, we would draw a table separating it into 1's, 10s, and 100's place writing out how many flats we have, and how many rods we have. In this case, we have 4 in the 1's place resembling 4 wholes, and, counting up the number of rods, we end up with 12 in the 10's place. Because there are no unit cubes the number in the 100's place is 0. For the 10's place, since we can't have more than one-digit numbers in each place, we would take the 1 from 12 and add it to 4 leaving us with 5 in the one's place and 2 in the ten's place.
Therefore, 4 x 1.3 = 5.2
Ex. 3 x 1.25 Using base 10 blocks, we have 3 flats, 2 rods under each flat, and 5 units under each group of 2 rods. Now, drawing a table with a 1's, 10's, and 100's place We have 3 in the 1's place, 6 in the 10's place since there is a total of 6 rods, and 15 in in the 100's place since there is a total of 15 units. Taking the 1 from the 15 and adding it to the 6 in tenths place, we end up with 3.75. Therefore, 3 x 1.25 = 3.75
Ex. 0.37 using base 10 blocks = 0 flats, 3 rods (0.3 or 3/10), and 7 units (0.07 or 7/100)
Ex. 0.8 using base ten blocks = 0 flats, and 8 rods (0.8 or 8/10)
Ex. 1.42 using base 10 blocks = 1 flat, 4 rods (0.4 or 4/10), and 2 units (0.02 or 2/100)
Ex. 3.0 using base 10 blocks = 3 flats, 0 rods, and 0 units
Ex. 16.51483 =
1x10 + 6x1+ 5/10^ (1) +1/10^ (2) + 4/10^ (3) + 8/10^ (4) + 3/10^ (5)
Since the 1 is in the tens place and the 6 is in the one's place before the decimal, we would multiply those numbers by their place values. After the decimal, we would divide each number by 10 adding an exponent to the 10 depending on how many digits there are after the decimal. The more digits after the decimal, the bigger the exponents will get.
Ex. 3.4452 3x1+ 4/10^ (1) + 4/10^ (2) + 5/10^ (3) + 2/10^ (4)
Ex. 51.003 fifty-one and three thousandths
Ex. 46.9 forty-six and nine tenths
Ex. 2.58 two and fifty-eight hundredths
Ex. 125.6427 one hundred twenty-five and six thousand, four hundred twenty-seven ten-thousandths.
Ex. 34.477 thirty-four and four hundred seventy-seven thousandths When we reach the decimal point, we always read it as "and".
Ex. |-3| = 3 Ex. |5| = 5 Ex. |x| = 6 In this particular example, x can be either 6 or -6 because when finding the absolute value between these numbers, the results would be the same. |6| = 6 and |-6| = 6.
Ex. |-7|+3 |-7| = 7, so 7 + 3 = 10 Ex. |x + 4| = 8 To solve this absolute value equation, we would first write this equation, then write a second equation of |x + 4| = -8. Next, we would solve each by subtracting 4 on both sides. The two 4's on either side cancel out and for the first equation, we would get 4 when subtracting 8 by 4. For the second equation, we would get -12 when subtracting -8 by 4. So, 4 and -12 are the two solutions for this equation. Ex. |1x - 3| = 2 To solve this absolute value equation, we would do the same thing for the last example, but now we would also divide by 1 for this particular example. So, when solving 1x - 3 = 2 , 2 + 3 = 5 divided by 1 which is 5. When solving 1x - 3 = -2, -2 + 3 = 1 divided by 1 which is 1. So, 5 and 1 are the two solutions for this equation.
Ex. |-7|+3
|-7| = 7, so 7 + 3 = 10
Ex. |x + 4| = 8
To solve this absolute value equation,
we would first write this equation, then write
a second equation of |x + 4| = -8. Next,
we would solve each by subtracting 4 on both sides.
The two 4's on either side cancel out and for the first
equation, we would get 4 when subtracting 8 by 4.
For the second equation, we would get -12 when subtracting
-8 by 4. So, 4 and -12 and the two solutions for this equation.
Ex. |1x - 3| = 2
To solve this absolute value equation, we would do the same thing for the last example, but now we would also divide by 1 for this particular example.
So, when solving 1x - 3 = 2 , 2 + 3 = 5 divided by 1 which is 5.
When solving 1x - 3 = -2, -2 + 3 = 1 divided by 1 which is 1.
So, 5 and 1 are the two solutions for this equation
Ex. |5x + 6| = -3 There is no solution for this absolute value equation since the result in this example is negative. Ex. 2|x - 1|+ 4 = 8 For this equation, we would subtract the 4 by itself and the 8 the two 4's cancel each other out leaving us with 8 - 4 which is 4. Next, we would divide 4 by 2 which gives us 2. Now that we've done all that, the new equation would be |x - 1| = 2. We can then write this equation and the second equation of x - 1 = -2. Then, add the 1 on both sides. the two 1's on either side cancel each other out and for the first equation, we get 3, and for the second equation, we get -1. So, 3 and -1 are the two solutions for this equation.
Ex. |5x + 6| = -3
There is no solution for this absolute value
equation since the result in this example is negative.
Ex. 2|x - 1|+ 4 = 8
For this equation, we would subtract the 4 by itself and the 8
the two 4's cancel each other out leaving us with 8 - 4 which is 4. Next, we would divide 4 by 2 which gives us 2. Now that we've done all that, the new equation would be |x - 1| = 2.
We can then write this equation and the second equation of
x - 1 = -2. Then, add the 1 on both sides. the two 1's on either side cancel each other out and for the first equation, we get 3,
and for the second equation, we get -1. So, 3 and -1 are the two solutions for this equation.
Absolute Value Video:
Since some of the yellow chips, which are positive, were taken away, the total value decreased compared to our original value.
Purple chips: (negative numbers) Green chips: (positive numbers) Let's say we start with -5 purple chips and 6 green chips. Adding -5 and 6 together gives us a total value of 1. Now say that 3 purple chips were taken away leaving us to subtract -2 by 6 which gives us a new value of -8.
Since some of the purple chips, which were negative, were taken away, the total value decreased compared to the original value.
Red chips: (negative numbers) Yellow chips: (positive numbers) Say we begin with -3 red chips and 2 yellow chips. Adding those together gives us a total value of -1. Now, let's say we are given 4 more red chips. Adding -1 by -4 gives us a new value of -5
Since some red chips, which were negative, were given instead of taken away, the total value still decreased compared to the original value
Purple chips: (negative numbers) Green chips: (positive numbers) Let's say that we begin with -6 purple chips and 4 green chips. Adding these two together gives us a total value of -2. Now say we are given 3 more green chips in which, this time, we would add -6 by 7 which gives us a new value of 1.
Since some green chips, which were positive were given, this time, the total value increased compared to the original value.
Subtopic
Equal Addends: This algorithm comes in handy when you're subtracting two numbers but some of the numbers on the top are smaller than the ones on the bottom. In this case, you would have to figure out how much to add on the top and bottom in order to have the bottom number have 0 in the one's places. For example 224 - 117. 4 is less than 7 so we could add 3 to the top and bottom numbers which would result in 227 - 120. Now we can easily subtract these two numbers. The result is 107.
Standard Form: The simplest way to subtract. Ex. 34 - 23 = 11, 526 - 314 = 212. Here are some examples when including base numbers. Ex. 58base 4 - 36base 4 = 22base 4, 743base 6 - 241base 6 = 502base 6
For example, 11010base 2 to base 10. From right to left, 0 = 2^0, 1 = 2^1, 0 = 2^2, 1 = 2^3, and 1 = 2^4. So, the equation would be: 1 x 16 + 1 x 8 + 0 x 4 + 1 x 2 + 0 x 1 which comes to a total of 26. So, 11010base 2 = 26base 10.
