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artcher [175]
1 year ago
6

13....................

Mathematics
1 answer:
Mashcka [7]1 year ago
6 0

13.

(a)

y=-5x^2+21x-3

for x = 2:

y=-5(2)^2+21(2)-3=-20+42-3=19

for x = 5

y=-5(5)^2+21(5)-3=-125+105-3=-23

(b)

The graph is:

(c)

Using the graph:

\begin{gathered} x=1.8,y=18.6 \\ x=0.548,y=7 \\ x=3.65,y=7 \end{gathered}\begin{gathered} x=1.8 \\ y=-5(1.8)^2+21(1.8)-3=-16.2+37.8-3=18.6 \end{gathered}\begin{gathered} y=7 \\ -5x^2+21x-3=7 \\ -5x^2+21x-10=0 \end{gathered}

Using the quadratic formula:

\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-21\pm\sqrt[]{21^2-4(-5)(-10)}}{2(-5)} \\ x=\frac{-21\pm\sqrt[]{241}}{-10} \\ x=0.54758 \\ or \\ x=3.6524 \end{gathered}

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If $8700 is invested at 3% annual simple interest, how much should be invested at 6% annual simple interest so that the total ye
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that is \bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}

it will yield some amount

subtract that amount from 393
the difference is how much the yield will be on the 6% investment
so

\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}
\\\\\\
\implies \boxed{?}\\\\
-----------------------------\\\\
\textit{how much to invest at 6\%?}
\\\\\\


\bf \qquad \textit{Simple Interest Earned}\\\\
(393-\boxed{?}) = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
t=years\to &1
\end{cases}
\\\\\\
\textit{solve for "P", to see how much should the Principal be}\\\\
\textit{keep in mind that }P+\boxed{?}=393\leftarrow \textit{both yields added}
6 0
3 years ago
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