13. Find the surface area of each figure to the nearest tenth.

Mathematics

1 answer:

mrs_skeptik [129]2 years ago

8 0

Check the picture below.

so, notice, the prism is really just 6 rectangles stacked up to each other at the edges.

front and back, two rectangles of 11x6

left and right, two rectangles of 5x6

top and bottom, two rectangles of 11x5

get the area of all 6, and sum them up, and that's the area of the prism.

$\bf \stackrel{\textit{front and back}}{2(11\cdot 6)}~~+~~\stackrel{\textit{left and right}}{2(5\cdot 6)}~~+~~\stackrel{\textit{top and bottom}}{2(11\cdot 5)}$

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During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th

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T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
P=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}$

by the third day, t = 3, there are 40 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &40\\
P=\textit{initial amount}\to &2\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r
\\\\\\
\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}$

how many folks are there sick by t = 6? $\bf \stackrel{that~many}{A=2(2.7)^6}$