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sleet_krkn [62]

4 years ago

5

Verify 2cotx=csc^2xsin2x

1 answer:

Lerok [7]4 years ago

4 0

$2\cot x=\csc^2x\sin2x\\\\R=\left(\dfrac{1}{\sin^2x}\right)(2\sin x\cos x)=\dfrac{2\sin x\cos x}{\sin^2x}=\dfrac{2\cos x}{\sin x}=2\left(\dfrac{\cos x}{\sin x}\right)=2\cot x\\\\L=2\cot x\\\\\boxed{L=R}\\\\Used:\\\\\csc x=\dfrac{1}{\sin x}\\\\\cot x=\dfrac{\csc x}{\sin x}$

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A minute hand of a clock is 16 in long. Find the distance traveled by the tip of the minute hand in one hour

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The length of a clock hand is the length of the radius of a circle.

The minute hand needs to travel a full circle in one hour.

This length of travel would be the circumference of the circle.

To find the circumference of a circle multiply the radius by 2 and then multiply by PI (π).

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Circumference = 32 x PI

Using 3.14 for PI:

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3 years ago

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3 years ago

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10

kompoz [17]

Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$

6 0

3 years ago