Take 4,000 times 12 to get 48,000. that’s how much A costs.

B is a little harder. you have to take (2,300+300)x12+(4*250). this equals $32,100

the correct answer would be B.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=3%5Csqrt%7B5%7D" id="TexFormula1" title="3\sqrt{5}" alt="3\sqrt{5}" align="absmiddle" class="l

adoni [48]

Answer:

3/5???

Step-by-step explanation:

6 0

2 years ago

The sum of 30 times (1/3)^(n-1) from 1 to infinity

sergeinik [125]

Let

$S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}$

Then

$\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}$

and

$S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}$

and as $n\to\infty$ , we end up with

$\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32$

So we have

$\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45$

8 0

3 years ago