HELP WITH THESE 4 QUESTIONS PLEASE ITS DUE IN A MINUTE

Someone plz give me a right answer now

olya-2409 [2.1K]

Answer:

$\sqrt{b {}^{2} + (c - a) {}^{2} }$

5 0

3 years ago

Could you please look at this question?

MrMuchimi

Answer:

a. see attached

b. H(t) = 12 -10cos(πt/10)

c. H(16) ≈ 8.91 m

Step-by-step explanation:

The cosine function has its extreme (positive) value when its argument is 0, so we like to use that function for circular motion problems that have an extreme value at t=0. The midline of the function needs to be adjusted upward from 0 to a value that is 2 m more than the 10 m radius. The amplitude of the function will be the 10 m radius. The period of the function is 20 seconds, so the cosine function will be scaled so that one full period is completed at t=20. That is, the argument of the cosine will be 2π(t/20) = πt/10.

The function describing the height will be ...

H(t) = 12 -10cos(πt/10)

The graph of it is attached.

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See part a.

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The wheel will reach the top of its travel after 1/2 of its period, or t=10. Then 6 seconds later is t=16.

H(16) = 12 -10cos(π(16/10) = 12 -10cos(1.6π) ≈ 12 -10(0.309017) ≈ 8.90983

The height of the rider 6 seconds after passing the top will be about 8.91 m.

8 0

3 years ago

in a class of 240 pupils, 2/5 of them are boys and the rest are girls. a) what is the fraction of girls? b) how many boys are in

Sergio [31]

A. 3/5 of the class are girls.
B. 96 of the pupils are boys.

A. To find this answer you subtract 2 from 5 to get 3 (numerator or top of fraction) then use 5 as the denominator or bottom of the fraction.

B. To find this answer, you divide 240 by 5, to show how many pupils are in 1/5 of the class (48). Then multiply 48 by 2, to get the 2/5 of the pupils who are boys (96).

I hope this helps!

7 0

3 years ago

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.

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