2 years ago

7. Gabriela deposits $500 in an ordinary annuity at the end of each month in an account earning 8% annual

Mathematics

1 answer:

bixtya [17]2 years ago

7 0

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3 over 4 open parentheses 1 half x space minus space 12 close parentheses space plus space 4 over 5

lesantik [10]

For this case we have the following expression:

$\frac {3} {4} (\frac {1} {2} x-12) + \frac {4} {5}$

Rewriting the expression we have:

We apply distributive property to the terms of parentheses:

$\frac {3} {8} x- \frac {36} {4} + \frac {4} {5}\\\frac {3} {8} x-9 + \frac {4} {5}\\\frac {3} {8} x- \frac {41} {5}\\$

Answer:

$\frac {3} {8} x- \frac {41} {5}$

4 0

3 years ago

Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$