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Strike441 [17]
2 years ago
10

Solve the exponential equation. 3(e^3)-14=11

Mathematics
1 answer:
MrMuchimi2 years ago
5 0

Answer:

2.027 to the nearest thousandth.

Step-by-step explanation:

3(e^3) - 14 = 11

3(e^3) =  11 + 14 = 25

e^3 = 25/3

e = ∛(25/3)

e = 2.027.

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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
2 years ago
Read 2 more answers
A student did a survey on how people feel when they are hungry. The results were that more people felt a stabbing pain than thei
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7 0
3 years ago
On average, Tony the turtle completes a 15-foot course in 1 minute, and Antonio the turtle completes an 18-foot course in 1 minu
marishachu [46]

Answer:

15 feet.

Step-by-step explanation:

Tony's speed = 15 ft/min.

Antonio's = 18 feet /min.

So in 5 minutes Tony travels 5*15 = 75 feet and Antonio travels 5*18 = 90 feet.

So the answer is 90-75 = 15 feet.

8 0
3 years ago
Parallel to 7x+5y=3, through (9,-9)
love history [14]
The answer should be -5x + 7y = -108

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3 years ago
A pound of chocolate costs 7 dollars. Joe buys p pounds. Write an equation to represent the total cost c that Joe pays.
pickupchik [31]

Answer:

Total cost, c = 7p

Step-by-step explanation:

1pound =$7

Joe buys p pounds,

p pounds = $7 * p

p pounds = 7p

3 0
3 years ago
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