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4 years ago

Please help, and show work thank you :)

1 answer:

3 0

Having a $75 added cost to the total price, we can presume they will never equal. It is easy to see if you make it algebraic:

$m=months\\t=total\\\\ProviderA => t = 39.95m\\ProviderB => t = 39.95m+75$

They will never equal each other.

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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

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

3 years ago

Read 2 more answers

a cubic box has a side length of 2 feet. How many of these boxes could fit inside a larger cubic box whose base has a perimeter

kap26 [50]

So if a cubic box has a side length of 2 ft in order to find the volume we need to cube it or multiply the length times 3 times. so it would be 2x2x2=8. that is the volume of the fiest cube. in order to find the volume of the larger cube first you need to know the length of one of the sides. the base has 4 sides to it that are all equal so you divide 24 by 4 and get 6. then to find the volume you multiply 6 three times since it is also a cubic cube so it would be 6x6x6=216. now you know the volume of each cube so you divide 216 by 8 to find how many times the volume of the small cube would go into the volume of the larger cube and get 27 times.

6 0

3 years ago

Timed test please help

insens350 [35]

Last option: − ∞ < y < 5

8 0

3 years ago

A.)4x-3y=6<br> B.)4x+3y=6<br> C.) 4x-3y=-6<br> D.)4x+3y=-6

natima [27]

C would be my guess here

4 0

3 years ago

SOME ONE HELP ME PLZ FAST it's already late

ololo11 [35]

Answer:

Step-by-step explanation:

6 0

3 years ago