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3 years ago

7

If you won the lottery today, and had a choice to receive all the money now, or get the money in the future, what is one big pro

blem with getting it in the future?
a.
they may change their mind and not pay you

b.
the lottery people may go broke, or the lottery may be shut down before you get it all

c.
the money you get in the future will actually buy more because of inflation

d.
if you get it in the future, you can't invest it now

1 answer:

iragen [17]3 years ago

3 0

Answer:

a,b,c,and d..?

Step-by-step explanation:

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3y + 1 - 2y = -3 - 3y

s2008m [1.1K]

Answer:

Step-by-step explanation:

3y + 1 - 2y = -3 -3y

4y = -4

y = -1

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3 years ago

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name the rational numbers that are not integers from the list below -11,0,57,8.07, sqrt 13, sqrt 4, 2 1/6, -8/45, 1.1919919991..

weqwewe [10]

Sqrt 13 and 1.1919919991... are irrational, meaning that they can't be described in a fraction of one integer over another, like 1/3, 45/44 or 57/107, these numbers are rational. Most irrationals are known constants like e or π, endless non-repeating decimals, or roots of non-perfect numbers like 13, 7, 5 or 2.

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3 years ago

Find the distance between the points L and D.<br><br>A)-2<br>B)0<br>C)1<br>D)2

Alisiya [41]

Answer:

C) 1

there is only one space between L and D

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3 years ago

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

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

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3 years ago

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PLEASE HELP ILL GIVE BRAINLIEST

cestrela7 [59]

Yea I wanna say you have the right answer

8 0

3 years ago