The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened. Movie A: y=-x^{2}+35x+100 Movie B: y=-5x+275 Where x is number of days since the movies opened. When is the attendance for each movie the same?

Suman Cole 2021-02-12 Answered
The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened.
Movie A: \(\displaystyle{y}=-{x}^{{{2}}}+{35}{x}+{100}\)
Movie B: \(\displaystyle{y}=-{5}{x}+{275}\)
Where x is number of days since the movies opened.
When is the attendance for each movie the same?

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Expert Answer

Arnold Odonnell
Answered 2021-02-13 Author has 8924 answers
Formula used:
\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
Calculation:
Substituting value of equation 1 in equation 2,
\(\displaystyle-{x}^{{{2}}}+{35}{x}+{100}=-{5}{x}+{275}\)
Shifting the value,
\(\displaystyle-{x}^{{{2}}}+{35}{x}+{100}=-{5}{x}+{275}={0}\)
On solving,
\(\displaystyle-{x}^{{{2}}}+{40}{x}-{175}={0}\)
Hence,
\(\displaystyle{a}=-{1},\)
\(\displaystyle{b}={40},\)
\(\displaystyle{c}=-{175},\)
Substituting the formula
\(\displaystyle{x}={\frac{{-{40}\pm\sqrt{{{40}^{{{2}}}-{4}{\left(-{1}\right)}{\left(-{175}\right)}}}}}{{{2}{\left(-{1}\right)}}}}\)
So,
\(\displaystyle{x}={\frac{{-{40}\pm\sqrt{{{1600}-{700}}}}}{{-{2}}}}\)
\(\displaystyle{x}={\frac{{-{40}\pm\sqrt{{{900}}}}}{{-{2}}}}\)
\(\displaystyle{x}={\frac{{-{40}\pm{30}}}{{-{2}}}}\)
\(\displaystyle{x}={\frac{{-{40}+{30}}}{{-{2}}}}\)
\(\displaystyle{x}={5}\)
Value of y, substituting in equation 2,
\(\displaystyle{y}=-{5}{\left({5}\right)}+{275}\)
\(\displaystyle{y}={250}\)
For -30,
\(\displaystyle{x}={\frac{{-{40}-{30}}}{{-{2}}}}\)
\(\displaystyle{x}={35}\)
Value of y, substituting in equation 2,
\(\displaystyle{y}=-{5}{\left({35}\right)}+{275}\)
\(\displaystyle{y}={100}\)
Hence, \(\displaystyle{x}={5},{35}\ \text{and}\ {y}={250},{100}\)
Hence, the attendance of two movies is same on day 5 and day 35
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