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MissTica
3 years ago
8

All the dimensions of a parallelogram were multiplied by the same factor. If the area of the parallelogram went from 3 square ce

ntimeters to 12 square centimeters, the dimensions of the parallelogram were multiplied by what?
Mathematics
2 answers:
postnew [5]3 years ago
3 0

Answer:

I was multiplied by 4.

Step-by-step explanation:

Because 3 times 1 is not 12

and 3 times 2 is not 12

and 3 times 3 is not 12

but only 3 time 4 is 12.

kari74 [83]3 years ago
3 0

Answer:

The answer is 2

Step-by-step explanation:

The dimensions of the parallelogram are most likely 3 and 1, so if you multiply both by 2 you get 6 and 2, which equal 12 when multiplied together

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Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Simplify completely 12 times x to the third power minus 4 times x to the 2nd power plus 8 times x all over negative 2 times x.
Alenkinab [10]
The answer is: -6x^2+2x-4
5 0
3 years ago
sum of two numbers is 39. the greater number is 7 more than the smaller number. ( set up an equation )
pochemuha
Let x be the unknown number

2x +7 = 39

2x+7-7 = 39-7

2x = 32

2x/2 = 32/2

x=16

You use the balance method to solve. First, you cancel the +7 by subtracting the 7 ON BOTH SIDES to balance the equation.. After that, you are left with 2x, where you will need to isolate the x by dividing by 2 ON BOTH SIDES to balance it.

Hope this helps
3 0
3 years ago
A and b are positive integers and a-b=2. Evaluate the following:<br> 9^1/2b/3^a
OlgaM077 [116]

Answer:

\frac{1}{ 9 }

Step-by-step explanation:

\huge \frac{ {9}^{ \frac{1}{2}b } }{ {3}^{a} }  \\  \\  =  \huge \frac{ {( {3}^{2} )}^{ \frac{1}{2}b } }{ {3}^{a} } \\  \\  =  \huge \frac{ { {3}}^{2 \times  \frac{1}{2}b } }{ {3}^{a} }  \\  \\  \huge =  \frac{ {3}^{b} }{ {3}^{a} }  \\  \\  \huge =  \frac{1}{ {3}^{a - b} }  \\  \\   \huge=  \frac{1}{ {3}^{2} } \\  ( \because \: a - b = 2)\\  \\  \huge =  \frac{1}{ 9 }

5 0
3 years ago
How do I solve this equation
Anestetic [448]
-x - y = 8
2x - y = -1

Ok, we are going to solve this in 2 parts.  First we have to solve for one of the variables in one of the equation in terms of the other variable.  I like to take the easiest equation first and try to avoid fractions, so let's use the first equation and solve for x.

-x - y = 8      add y to each side
-x = 8 + y      divide by -1
x = -8 - y

So now we have a value for x in terms of y that we can use to substitute into the other equation.  In the other equation we are going to put -8 - y in place of the x.

2x - y = -1
2(-8 - y) - y = -1      multiply the 2 through the parentheses
-16 - 2y - y = -1      combine like terms
-16 - 3y = -1            add 16 to both sides
-3y = 15                   divide each side by -3
y = -5

Now we have a value for y.  We need to plug it into either of the original equations then solve for x.  I usually choose the most simple equation.

-x - y = 8
-x - (-5) = 8            multiply -1 through the parentheses
-x + 5 = 8                subtract 5 from each side
-x = 3                      divide each side by -1
x = -3

So our solution set is

(-3, -5)

That is the point on the grid where the 2 equations are equal, so that is the place where they intersect.

4 0
3 years ago
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