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aivan3 [116]
2 years ago
15

The triangle and square have equal perimeters. What is the perimeter of the triangle?

Mathematics
1 answer:
Bezzdna [24]2 years ago
6 0

Answer:

E. 16

Step-by-step explanation:

Perimeter of the triangle = 4x+4x+(x+7)

Perimeter of square = 4(x+3)

We know that the perimeters are equal so we can form an equation to solve for x:

9x+7 = 4x + 12

Now we can subtract 4x from both sides to move the xs to the left side, and the same with the 7 so we have the variable on one side and the constant on the other:

9x+7-7-4x=12+4x-4x-7

5x = 5

Now we divide both sides by 5 to get:

5x/5 = 5/5

x = 1

Now we can use this value to figure out the perimeter of the triangle:

P = 9x + 7

x=1, substitute into equation:

P = 9(1) + 7

P = 16

Hope this helped!

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Answer:

if the 3 is an exponent, then it is 4 bc 4 to the power of 3 is 64

Step-by-step explanation:

x3=64

Step 1: Take cube root.

x=(64)(13)

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Answer:

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3 years ago
Match each equation with its solution set.
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Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

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∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

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∴ x + 3 = -12 ⇒ subtract 3 from both sides

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* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Given :

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