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3 years ago

11

If 2x+5= 12 for some value of x, then what does 6x+20 equal???

2 answers:

ANEK [815]3 years ago

6 0

1. Solve for x in 2x + 5 = 12
2x + 5 = 12 Subtract 5 from both sides.
2x = 7 Divide 2 from both sides.
x = 3.5

2. Substitute 3.5 in for x in 6x + 20
6(3.5) + 20
21 + 20
41

Your answer is 41.

Morgarella [4.7K]3 years ago

5 0

Here is the answer to the question

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Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

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see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

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Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

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