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2 years ago

A construction job requires 16 pieces of steel, each 7.96 ft long. What length of steel is needed?

Mathematics

2 answers:

crimeas [40]2 years ago

4 0

16*7.96=127.36

127.36ft of steel will be required.

amm18122 years ago

4 0

16*7.96ft = 127.36ft

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9. Which of the following could be the side measures of a right triangle?

Lapatulllka [165]

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A. 5 ft, 4 ft, 3 ft

Step-by-step explanation:

When you use the Pythagorean theorem with the numbers of the first choice, you find that these could be the sides of a right triangle:

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If you want to check, you can examine the other choices:

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<em>Comment on 3-4-5 triangle</em>

As you can see from the above, side measures of 3, 4, and 5 will form a right triangle. This is the smallest set of integers that will do so. It is also the ratios of the only set of side lengths in arithmetic progression that will do so.

This latter fact can help you rule out choices B and D in this problem, because each of those has sides in the progression 1 : 2 : 3.

(3, 4, 5) triangles come up often in algebra and geometry problems. This is a useful "Pythagorean triple" to remember.

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3 years ago

Identify the equation for the graph shown.

mario62 [17]

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2 years ago

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

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3 years ago