Sonias final math exam had 174 questions. the number of algebra questions to the number of trigonometry questions is 14: 15. how many trigonometry questions were on the final exam?

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maybe you can attach the photo ?

f(g(x))  You plug in g(x) for x


f(x) = 2x + 3

f(g(x)) = 2(g(x)) + 3

f(g(x)) = 2(4x - 5) + 3

f(g(x)) = 8x - 10 + 3

f(g(x)) = 8x - 7


g(f(x))  You plug in f(x) for x


g(x) = 4x - 5

g(f(x)) = 4(f(x)) - 5

g(f(x)) = 4(2x + 3) - 5

g(f(x)) = 8x + 12 - 5

g(f(x)) = 8x + 7

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi<x<\dfrac{3\pi}2, we should expect \cos x<0. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi<x<\dfrac{3\pi}2\implies\dfrac\pi2<\dfrac x2<\dfrac{3\pi}4

so we should expect \tan\dfrac x2<0 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

45

Step-by-step explanation:

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step-by-step explanation:

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