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Masja [62]
3 years ago
9

PLEASE HELP What is the y intercept of this line?

Mathematics
2 answers:
Natasha_Volkova [10]3 years ago
8 0

Answer:

I'm pretty positive its (0, 0)

schepotkina [342]3 years ago
8 0
The answer has got to be (0,0)
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One eighth divided by five equals five eighths.
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Select the correct answer from each drop-down menu
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Step-by-step explanation:

uv =5

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wu = (5²+6²)½

wu = 7.8 approximate 8

perimeter = 6+5+8 = 19

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4x-5=6x+13<br> What is the first step in solving equations
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[Pre-Calc] Please Help! I don’t know where to start. How do I do this?
sertanlavr [38]

Answer:

See Below.

Step-by-step explanation:

Problem A)

We have:

\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta

When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.

So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:

\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta

Cancel:

\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta

Let 1/cosθ = secθ:

\sec^2\theta -1=\tan^2\theta

From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:

\tan^2\theta =\tan^2\theta

Problem B)

We have:

\sin^3x=\sin x-\sin x \cos^2 x

Factor out a sine:

\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x

From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:

\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x

Distribute:

\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x

Problem C)

We have:

\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x

Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:

\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x

From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:

\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x

Cancel:

\displaystyle \frac{\cos x}{\sin x}=\cot x

By definition:

\cot x = \cot x

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