<img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7Bx%2Bcosx%7D%7Bx%7D%20" id="TexFormula1" ti

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3 years ago

tle=" \lim_{x \to \infty} \frac{x+cosx}{x} " alt=" \lim_{x \to \infty} \frac{x+cosx}{x} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

grandymaker [24]3 years ago

6 0

The answer is 1.
\mathrm{Divide\:by\:highest\:denominator\:power:}\:1+\frac{\cos \left(x\right)}{x}

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Answer:

(a) $A(n)=5500(1.01)^{4n}$

(b)$5955.71

Step-by-step explanation:

For an initial principal P deposited in an account at an annual interest r compounded for a number of period k, the amount in the account after n years is given by the model:

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

(a)Aunt Ga Ga gave you $5,500 to save for college.

P=$5,500

Annual Interest, r=4%=0.04

Since interest is compounded quarterly, Number of Periods, k=4

Therefore, an exponential function modeling this situation is:

$A(n)=5500(1+\dfrac{0.04}{4})^{4n}\\A(n)=5500(1+0.01)^{4n}\\$Simplified\\A(n)=5500(1.01)^{4n}$

(b)After 2 years, i.e. when n=2

$A(2)=5500(1.01)^{4*2}\\=\$5955.71$

(c)When A(n)=$10000, we have:

$10000=5500(1.01)^{4n}\\$Divide both sides by 5500\\(1.01)^{4n}=\dfrac{10000}{5500} \\$To solve for n, we change to logarithm form\\Log_{1.01}\dfrac{10000}{5500}=4n\\= \dfrac{ Log \dfrac{10000}{5500}}{Log 1.01}=4n\\4n=60.08\\n=60.08 \div 4\\n=15.02\\$Therefore, in 15.02 years, the account would be worth $10,000.$