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3 years ago

15

Missing angle trig (inverse) 10th geometry HELP ASAP

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

33. B

34. D

35. D

36. A

Step-by-step explanation:

All these questions require is a calculator :). For question 35, all you want to put in is arccos(0.8192) and make sure your calculator is in degree mode. For 36, all you want to put in is arctan(0.3839). For 34, all you want to put in is arcsin(0.6018). For 33, all you want to put in is arcsin(0.6561).

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Answer:

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Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

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$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

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$=\frac{-2x-12}{(x+4)(x-4)}$

$=\frac{-2(x+6)}{(x+4)(x-4)}$

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

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