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Anestetic [448]

3 years ago

13

3. Mr. Gengel wants to make a shelf with boards that are feet long. If he has an -foot board, how many pieces can he cut from th

e big board?

1 answer:

ikadub [295]3 years ago

7 0

Sorry I dont know... if I'm late on this

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Sum of 2 numbers is 42, and their difference is 8. What are the numbers?

Tems11 [23]

Answer:

25 , 17

Step-by-step explanation:

Let the unknown two numbers be x & y.

According to the question,

Sum of 2 numbers is 42.

x + y = 42 ⇒ ( 1 )

Their difference is 8.

x - y = 8 ⇒ ( 2 )

First let us find the value of x.

( 1 ) + ( 2 )

x + y + x - y = 42 + 8

2x = 50

Divide both sides by 2.

x = 25

And now let us find the value of y.

x + y = 42

25 + y = 42

y = 42 - 25

y = 17

Therefore, the two numbers are 25 , 17

Hope this helps you :-)

Let me know if you have any other questions :-)

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3 years ago

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look at this equation and the steps used to solve it. Which answer is the best justification for step 1 of this solution

Julli [10]

Answer:

Step-by-step explanation:

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3 years ago

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sattari [20]

Answer:

5 x + 3 = 23

Step-by-step explanation:

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3 years ago

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Find: 11<br> 11/3<br> ÷ 2<br> 2/3<br><br> The quotient is 5 and __.

Zinaida [17]

Answer: The quotient is 5.5

Step-by-step explanation

4 0

3 years ago

Help! Prove the equality<br><br>arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

4 years ago