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kondor19780726 [428]

2 years ago

9

31-inch piece of steel is cut into three pieces so that the second piece is twice as long as the first piece and the third piece

is one inch more than seven times the length of the first piece, find the lengths of the pieces

1 answer:

pychu [463]2 years ago

4 0

"31-inch piece of steel is cut into three pieces"

Call the pieces a,b,c

a + b + c = 31

"second piece is twice as long as the first piece"

b = 2a

"third piece is one inch more than seven times the length of the first piece"

c = 1 + 7a

Ok, we have our equations, let's substitute the second and third into the first.

a + 2a + 1 + 7a = 31

10a + 1 = 31

10a = 30

a = 30/10 = 3

b = 2a = 6

c = 1 + 7a = 22

Check: 3 + 6 + 22 = 31, good

Answer: 3 inches, 6 inches, 22 inches

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Step-by-step explanation:

Use the 30°-60°-90° formulas:

a. $longer$ $leg=\sqrt{3}*shorter$ $leg$

b. $hypotenuse=2*shorter$ $leg$

1.) Insert values for a:

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Simplify:

$x=8\sqrt{3}$

Insert values for b:

$y=2*8$

Simplify:

$y=16$

2.) Insert values for a:

$y=\sqrt{3}*\frac{1}{2}$

Simplify:

$y=\frac{\sqrt{3}}{2}$

Insert values for b:

$x=2*\frac{1}{2}$

Simplify:

$x=1$

3.) Insert values for a:

$y=\sqrt{3}*14$

Simplify:

$y=14\sqrt{3}$

Insert values for b:

$x=2*14$

Simplify:

$x=28$

4.)Insert values for a:

$y=\sqrt{3}*12$

Simplify:

$y=12\sqrt{3}$

Insert values for b:

$x=2*12$

Simplify:

$x=24$

5.) Insert values for a:

$12=\sqrt{3}*x$

Divide both sides by $\sqrt{3}$ and rationalize:

$\frac{12}{\sqrt{3}}=\frac{\sqrt{3}*x}{\sqrt{3}}\\\\\frac{12}{\sqrt{3}}=x\\\\\frac{\sqrt{3}}{\sqrt{3}}*\frac{12}{\sqrt{3}}\\\\\frac{12\sqrt{3}}{\sqrt{9}}\\\\\frac{12\sqrt{3}}{3}\\\\4\sqrt{3}=x$

Flip:

$x=4\sqrt{3}$

Insert values for b:

$y=2*4\sqrt{3}$

Simplify:

$y=8\sqrt{3}$

6.) Insert values for a:

$8=\sqrt{3}*x$

Divide both sides by $\sqrt{3}$ and rationalize:

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Flip:

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Insert values for b:

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Simplify:

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