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LenKa [72]
3 years ago
12

Evaluate the expression when y = 3 B.) 12y

Mathematics
1 answer:
Archy [21]3 years ago
3 0
12y = 12*3 = 36  ← answer
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Braden logged 3 2/3 hours last week in his reading journal and 2 3/6 hours last week how many fewer hours did Brayden log in his
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1 1/6

Step-by-step explanation:

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3 years ago
What is 26% of 160?<br> A. 4.16<br> B. 6.15<br> C. 41.6<br> D. 61.5
Vinvika [58]

Answer:

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Step-by-step explanation:

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Read 2 more answers
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
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