The sum of the quotient of a number and and 3 and 8 is 16 what is the number

1 answer:

5 0

After having done this on paper:

$\frac{n}{3} + 8 = 16$

We want to isolate the variable, and the first step is to subtract the 8.

$\frac{n}{3} = 8$

Since the n is being multiplied by 3, and that = 8, we can multiply both sides by 3.

n = 24

Let me know if this helped you understand better.

Also, you can check this by substituting in 24 for n.

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5. Amy deposited $7,000 in a bank account earning 3.5% interest, compounded

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Answer:

Step-by-step explanation:

Given the following data;

Principal = $7,000

Interest = 3.5% = 3.5/100 = 0.035

To find the future value, we would use the compound interest formula;

$A = P(1 + \frac{r}{n})^{nt}$

Where;

A is the future value.

P is the principal or starting amount.

r is annual interest rate.

n is the number of times the interest is compounded in a year.

t is the number of years for the compound interest.

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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.