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3 years ago

8

Question is attached!

1 answer:

nadezda [96]3 years ago

6 0

Answer:

the answer to the question would be 25

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Write the standard form of the equation of each line.<br><br> 1. y= -2/5x+1<br> 2. y+3 = -3/2(x+1)

gavmur [86]

Answer:

1. 2x + 5y = 5

2. 3x + 2y = -9

Step-by-step explanation:

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2 years ago

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Josh book has 64 pages. Josh read 27 pages in the morning He reads 18 more pages that night How many pages does josh need to rea

Ainat [17]

Josh needs to read 19 more pages

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3 years ago

Karen claps her hand and hears the echo from a distant wall 0.113 s later. How far away is the wall? The speed of sound in air i

insens350 [35]

Answer:

19.3795 m

Step-by-step explanation:

If sound travels at 343 m/s and it took Karen 0.113 s to hear the echo from the wall, the distance travelled by the sound is:

$D=343\frac{m}{s} *0.113\ s\\D= 38.759\ m$

Note that the distance calculated above is the distance travelled from Karen to the wall and then back from the wall to Karen. Therefore, the distance between Karen and the wall is:

$d=\frac{38.759}{2}\\d=19.3795\ m$

The wall is 19.3795 m away from Karen.

3 0

3 years ago

Solve the proportion.<br><br> 2/17=x/68<br><br> A. 8<br><br> B. 34<br><br> C. 4<br><br> D. 578

Valentin [98]

Answer is A bc ur dividing the quotients together

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3 years ago

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Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

3 years ago