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3 years ago

Solve the math problem please

Mathematics

2 answers:

DedPeter [7]3 years ago

6 0

Answer:

Choice B

Step-by-step explanation:

TRUST ME. IF YOU WANT EXPLANATION:

54 - 10x < = 20 + 7x

54 - 20 < = 7x + 10x

34 < = 17x

34/17 < = x

2 < = x.....or x > = 2....its B trust me

mars1129 [50]3 years ago

5 0

Answer:

B, x≥2

Step-by-step explanation:

54 - 10x ≤ 20 + 7x - subtract 20

34 - 10x ≤ 7x - add 10x to 7x

34 ≤ 17x - divide 34 by 7

2 ≤ x - flip the equation - x ≥ 2

Hope this is helpful!

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$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
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then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

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