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3 years ago

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a boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours. if b=the rate of the boat in still wate

r and c=the rate of the current, which of the following equations represents the trip upstream?

1 answer:

ASHA 777 [7]3 years ago

8 0

it would be equation c, since you are you going upstream and the current would be going down stream

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3 years ago

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3 years ago

Read 2 more answers

Evaluate the definite integral from pi/2 to put of cos theta/sqrt 1+ sin theta.

masha68 [24]

Answer:

$\textsf{B.}\quad -2(\sqrt{2}-1)$

Step-by-step explanation:

Given integral:

$\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta$

Solve by using <u>Integration by Substitution</u>

<u />

Substitute u for one of the functions of $\theta$ to give a function that's easier to integrate.

$\textsf{Let }u=1+\sin \theta$

Find the derivative of u and rewrite it so that $d \theta$ is on its own:

$\implies \dfrac{du}{d \theta}=\cos \theta$

$\implies d \theta=\dfrac{1}{\cos \theta}\:du$

Use the substitution to change the limits of the integral from $\theta$ -values to u-values:

$\textsf{When }\theta=\pi \implies u=1$

$\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2$

Substitute everything into the original integral and solve:

$\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}$

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1 year ago

Soloha48 [4]

They did you dirty:((((

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2 years ago