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charle [14.2K]
3 years ago
12

ZEARN MATH

Mathematics
1 answer:
joja [24]3 years ago
3 0

Answer:

ok ill do it :-)

Step-by-step explanation:

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Answer:

B.. ..... .., .........

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The perimeter of a geometric figure is the sum of the lengths of its sides. If the perimeter of the pentagon to the right (five-
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2.5 divided by 2 equals 1.25 so each side equals 1.25
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Solve this system of linear equations. Separate
alex41 [277]

Answer:

y = 7 and x =3

Step-by-step explanation:

use elimination method

so we gonna eliminate y first

9x = 1 - 4y

-7x = -7 + 4y

7| 9x = 1 - 4y

9| -7x = -7 + 4y

63x = 7 - 28y

-63x = -63 + 36y

add eqtn 1 to eqtn2

you will get

0 = -56 + 8y

56 = -56 + 56 + 8y

56 = 8y

56÷8 = 8y÷ 8

7 = y

also eliminate x as we have done wth y

4| 9x = 1 - 4y

4| -7x = -7 + 4y

36x = 4 - 16y

-28x = -28 + 16y

add the two equations

u will get

8x = -24 + 0

8x = -24

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5 0
2 years ago
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
2 years ago
If 4^x+1 = 64, then what is the value of x^4
Olin [163]
4^{x} + 1 = 64 \\4^{x} = 63 \\ln(4^{x}) = ln(63) \\xln(4) = ln(63) \\x = \frac{ln(63)}{ln(4)}

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