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ehidna [41]
3 years ago
5

Please help this is due and i only got the second part done

Mathematics
1 answer:
Ainat [17]3 years ago
8 0

Answer:2n+13=75 n=31

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Find the illegal values of b in the fraction (2b^2+36-10)/(b^2-26-8)
skad [1K]
Correct Question is: Find the illegal values of b in the fraction (2b^2 + 3b - 10)/(b^2 - 2b - 8)

illegal values are those which are not a part of Domain. In this case these will be such values which will make the denominator of the fraction zero.

So setting the denominator equal to zero, we can find these values.

b^{2} -2b-8=0 \\  \\ 
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b(b-4)+2(b-4)=0 \\  \\ 
(b+2)(b-4)=0 \\  \\ 
b=-2, b=4

Thus, b=-2 and b=4 are the illegal values for this expression
6 0
4 years ago
What number should be added to each expression to make a perfect square trinomial?
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2x+3x+4
ivolga24 [154]
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3 years ago
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is it possible to have a right triangle in which the sine of one of the angles is equal to the cosine of one of the angles? why?
stira [4]

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Yes , it is possible (see the explanation)

Step-by-step explanation:

we know that

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so

If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa

<em>Example</em>

Suppose we have the right triangle ABC

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A+B=90°

therefore

sin(A)=cos(B)

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zavuch27 [327]

Step-by-step explanation:

Hey there!

See explanation in picture.

<em><u>Hope </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

7 0
3 years ago
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