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3 years ago

Find the distance traveled at the given speed and time using d=rt and r= 32mph and t= 4.5hr.

2 answers:

6 0

We are told that r = 32 mph and t = 4.5 hrs and that d = rt. r and t are together by multiplying. Thus, d = rt = (32)(4.5) = 144 miles.

We use miles as the unit because the hours from the rate in the simplify to one with the hours in the time.

So d = 144 miles.

Goshia [24]3 years ago

3 0

Hello!

As you can see, our distance is the product of our rate and time. Let's multiply below.

32(4.5)=144

Therefore, you would have traveled 144 miles.

I hope this helps!

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Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

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Please answer the questions down below. Savvas Realize 10-1 Interactive Additional Practice.

masha68 [24]

9. If they come in packages of 8 he needs 12 1/2

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1 year ago

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Rus_ich [418]

3/5 / 3/7

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3 years ago

A club began with 3 members. Each month, each member brought one new member. Which function can be used to determine the number

erastova [34]

In x th month, members = $3^{2x}$ . Option D.

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Given,

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To find the number of members after x months.

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In 1st month, members = 3

In 2nd month, members = 3×3 = 9 = 3²

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Hence,

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3 years ago

What is the x-intercept of the line containing the points (–6, 10) and (12, –2)?

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equation
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2 years ago

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