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diamong [38]
3 years ago
13

Find S15 for the geometric series 72 + 12 + 2 + 1/3 +

Mathematics
2 answers:
Rama09 [41]3 years ago
8 0
Finish the equation. please and thank you

sweet-ann [11.9K]3 years ago
5 0
Hello,

u_{1}=72\\

u_{2}=\dfrac{72}{6}\\

u_{3}=\dfrac{72}{6^{2}}\\

u_{15}=\dfrac{72}{6^{14}}\\


s=72*(1+\dfrac{1}{6^{1}}+\dfrac{1}{6^{2}}+\dfrac{1}{6^{3}}++...+\dfrac{1}{6^{14}})\\
=72*(\dfrac{\frac{1}{6^{15}}-1}{\frac{1}{6}-1}})\\
=72*(\dfrac{470184984576-1}{78364164095*5})\\

=86,400...



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Derivative of tan(2x+3) using first principle
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The derivative is given by the limit

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Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
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\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

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3 years ago
Your housing and fixed expenses are $702.38 per month do you want to have a four month emergency fund and save it over a nine mo
MrMuchimi

Answer:

For Review: Planning Ahead and Contracts Quick Check... these are the answers...

1) B. Yes, you want to begin saving at least 25 years before you plan to retire

2) D. all of the above

3) $2,341.27 per month realized income

4) Your housing and fixed expenses are $702.38 per month do you want to have a four month emergency fund and save it over a nine month period of time how much do you need to save per month?

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Answer) $312.17 per month, for 9 months to save 4 months worth or expenses.

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