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3 years ago

the cost of six months of spotify is 59.70. if there is a $5 monthly fee what is the cost of one spotify

Mathematics

1 answer:

Naddik [55]3 years ago

6 0

Answer:

14.95

Step-by-step explanation:

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If $17,800 is invested at an interest rate of 8% per year, find the value of the investment at the end of 5 years for the follow

marta [7]

Answer:

7120

Step-by-step explanation:

6 0

3 years ago

The amount of radioactive uranium changes with time. The abscissas and ordinates of the points given are the time and amoung of

kakasveta [241]

The answer is A. f(t) = 100(0.25)t

This may be solved by inserting all points to your answer and sseeing if they work. Since A. f(t) = 100(0.25)t fits with all points, this best represents the relationship between f(t) and t.

Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.

4 0

3 years ago

This question is confusing. Please help me!

patriot [66]

Answer:

C. 490.

Step-by-step explanation:

25% = 0.25.

If the number of emails that Ryan and Taylor each received is x emails then Sara received 2x + 0.25 * 2x = 2.5 emails.

So 2x + 2.5x = 882

x = 882/4.5

x = 196.

So Sara received 2.5 * 196

= 490 emails.

5 0

3 years ago

5. Find the 34th term in the arithmetic sequence 9, 6, 3, A. 90 B. -90 C.-57 D. 108

mafiozo [28]

Answer:

B. -90

Step-by-step explanation:

a1=9; d=-3

a34=9-3(33-1)=-90

8 0

3 years ago

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago