Rachel earns $42,000 per year. Find how much she would make if she was paid weekly, semimonthly, and monthly.

Mathematics

2 answers:

liq [111]2 years ago

5 0

Answer:

Below! : )

Step-by-step explanation:

$3,500 monthly

$807.69 per week

$1,615.38 biweekly

Hope this helps! : )

Delicious77 [7]2 years ago

4 0

Answer:

in a monthly shy much $3500

in a weekly she much $807.69

Step-by-step explanation:

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For what values of θ is sinθ < cosθ when 0 ≤ θ < π ?

zysi [14]

<h3>Answer: $0 \le \theta < \frac{\pi}{4}$ </h3>

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How to get this answer:

Use the unit circle to note that $\sin\theta = \cos\theta = \frac{\sqrt{2}}{2}$ when $\theta = \frac{\pi}{4}$ (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$

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Here's a more detailed picture of whats going on.

$\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\$