$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$1400\\ P=\textit{original amount deposited}\dotfill &\$1100\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &7 \end{cases}$

$1400=1100\left( 1+\frac{r}{12} \right)^{12\cdot 7}\implies \cfrac{1400}{1100}=\left( 1+\frac{r}{12} \right)^{84}\implies \cfrac{14}{11}=\left( 1+\frac{r}{12} \right)^{84} \\\\\\ \sqrt[84]{\cfrac{14}{11}}=1+\cfrac{r}{12}\implies \sqrt[84]{\cfrac{14}{11}}-1=\cfrac{r}{12}\implies 12\left[ \sqrt[84]{\cfrac{14}{11}}-1 \right]=r \\\\\\ 0.0345\approx r\stackrel{\textit{converting to \%}}{0.0345\cdot 100}\implies 3.45\approx r\implies \stackrel{\textit{rounded up}}{3.4\approx r}$

8 0

3 years ago

Q3.

Xelga [282]

Answer:

x = 12

Step-by-step explanation:

Given x varies as the cube root of y then the equation relating them is

x = k $\sqrt[3]{y}$ ← k is the constant of variation

To find k use the condition x = 6 when y = 8, thus

k = $\frac{x}{\sqrt[3]{y} }$ = $\frac{6}{\sqrt[3]{8} }$ = $\frac{6}{2}$ = 3

x = 3 $\sqrt[3]{y}$ ← equation of variation

When y = 64, then

x = 3 × $\sqrt[3]{64}$ = 3 × 4 = 12

6 0

3 years ago

Subtract 8 – 63.

Vsevolod [243]

Answer:

-55

Step-by-step explanation:

4 0

4 years ago

What is the surface area of the right prism?

Mama L [17]

2 surfaces are 6 in by 9 in, meaning that their combined area is 2(54 in^2).
2 surfaces are 6 in by 18 in; combined area is 2(108 in^2).
2 surfaces are 9 in by 18 in; combined area is 2(162 in^2).

Add up these areas: A = 2(54+108+162) in^2 = 648 in^2 (answer)

4 0

3 years ago