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Lelu [443]
3 years ago
10

The attendance at a team's basketball game can be approximated with the polynomial-5x^2+80x+285, where x is the number of wins t

he team had in previous months.
Factor the polynomial completely.

Estimate the attendance if the teams won 4 games in the previous month.
Mathematics
1 answer:
dalvyx [7]3 years ago
8 0
That would be 525. :)
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For what value of a should you solve the system of elimination?
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\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}

\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20
\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12

\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}

3a-10y=-8 \ \textgreater \  \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}
3a-10y-3a=-8-3a

\mathrm{Simplify} \ \textgreater \  -10y=-8-3a \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}-10
\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}

Simplify more.

\frac{-10y}{-10} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \  \frac{10y}{10}

\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \  y

-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{-8-3a}{-10}

\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \  -\frac{-3a-8}{10} \ \textgreater \  y=-\frac{-8-3a}{10}

\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \  6x+10\cdot \frac{8}{10-3a}=20

10\cdot \frac{8}{10-3a} \ \textgreater \  \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{8\cdot \:10}{10-3a}
\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \  \frac{80}{10-3a}

6x+\frac{80}{10-3a}=20 \ \textgreater \  \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}
6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}

\mathrm{Simplify} \ \textgreater \  6x=20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \  \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}

\frac{6x}{6} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \  x

\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{20-\frac{80}{-3a+10}}{6}

20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \  \frac{20}{1}-\frac{80}{-3a+10}

\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10

Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \  \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}
\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \  \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}

20\left(-3a+10\right)-80 \ \textgreater \  Rewrite \ \textgreater \  20+10-3a-4\cdot \:20

\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \  20\left(-3a+10-4\right) \ \textgreater \  Factor\;more

10-3a-4 \ \textgreater \  \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \  -3a+6 \ \textgreater \  Rewrite
-3a+2\cdot \:3

\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \  3\left(-a+2\right) \ \textgreater \  3\cdot \:20\left(-a+2\right) \ \textgreater \  Refine
60\left(-a+2\right)

\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \  \frac{10\left(-a+2\right)}{\left(-3a+10\right)}

\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \   \frac{10\left(-a+2\right)}{-3a+10}

Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}

Hope this helps!
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