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2 years ago

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Writing a recursive formula..

2 answers:

OLEGan [10]2 years ago

5 0

$a_1=\frac{24}{2^1}+4=16\\\\a_2=\frac{24}{2^2}+4=10\\\\a_3=\frac{24}{2^3}+4=7\\\\a_4=\frac{24}{2^4}+4=5.5\\\\a_5=\frac{24}{2^5}+4=4.75\\\\a_6=\frac{24}{2^6}+4=4.375\\\vdots\\a_n=\frac{24}{2^n}+4$

wariber [46]2 years ago

3 0

A1=21+24/2!=16

Hope I helped

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5q - p + p + 1, please help!

Mamont248 [21]

So basically you just add the like terms to simplify the equation. You have 5q-p+p+1 you can cancel out the both p’s because one is negative and one is positive. That leaves you with 5q + 1 they are not like terms so you cannot simplify them so your simplified equation is 5q +1

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2 years ago

If Free Yer Assets Bank (FYAB) will give you 0.25% compounded quarterly and I.M.A.Q.T.π bank will give you 0.23% interest compou

ivann1987 [24]

Answer:

FYAB gives a better deal.

Step-by-step explanation:

Compound interest:

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

Continuously compounded interest:

$A = Pe^{rt}$

For the quarterly compounded interest, r = 0.25%, and n = 4.

$2P = P(1 + \dfrac{0.0025}{4})^{4t}$

$1.000625^{4t} = 2$

$\log (1.000625^{4t}) = \log 2$

$4t(\log 1.000625) = \log 2$

$t = \dfrac{\log 2}{4\log 1.000625}$

$t = 277$

For the continuously compounded interest, r =0.23%

$A = Pe^{rt}$

$2P = Pe^{0.0023t}$

$2 = e^{0.0023t}$

$\ln 2 = \ln e^{0.0023t}$

$\ln 2 = 0.0023t$

$t = \dfrac {\ln 2}{0.0023}$

$t = 301$

The quarterly compounded doubles in 277 years.

The continuously compounded doubles in 301 years.

Answer: FYAB gives a better deal.

7 0

2 years ago

Solve the equation v+5/-9 - 4 = -6/8

tamaranim1 [39]

First you simplify both sides of the equation to make v+-41/9=-3/4.

Then add -41/9 to both sides to get the answer which is 137/36

v=137/36

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Answer:

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Step-by-step explanation:

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