Respuesta :
The two lines intersect at the point (4, 214). This means that the attendance was 214 for both plays on the 4th night.
B. The attendance was the same on day 4. The attendance was 214 at both plays that day.
B. The attendance was the same on day 4. The attendance was 214 at both plays that day.
Option A and B are correct. That is , "the attendance was same on day 40 and attendance was 790" and " the attendance was same on day 4 and the attendance was 214".
Solving systems by substitution:
The method of substitution involves three steps :
- Solve on equation for one of the variable.
- substitute this expression into the other equation and solve.
- Resubstitute the value into the original equation to find the corresponding variable.
According to the given question
We have two equations
A: [tex]y=16x+150[/tex]
B: [tex]y=-x^{2} +60x-10[/tex]
where, x represents "the number of days"
and, y represents "the number of attendance".
For finding the same day(s) when the attendance was same at both plays, we have to solve equations A and B for x and y.
since, [tex]y = 16x+150\\[/tex]
Substitute the value of y in equation B.
Therefore, we get a new equation
[tex]16x+150 =-x^{2} +60x-10[/tex]
⇒[tex]x^{2} +16x -60x+150+10=0\\x^{2} -44x+160=0[/tex]
Solve, the above quadratic equation by quadratic formula
x = {-(-44) ± [tex]\sqrt{(-44)^{2}-4(160) }[/tex]}/2
x = (44± [tex]\sqrt{1936-640}[/tex])/2
x = [tex]\frac{44}{2}[/tex] ±[tex]\frac{\sqrt{1296} }{2}[/tex]
x = 22 ± [tex]\frac{36}{2}[/tex]
x = 22 ± 18
x = 22+18 and 22-18
x = 40 and 4
When, x=4
y = 16(4) +150 =64+150 = 214
When, x = 40
y = 16(40) + 150 = 640 + 150 = 790
So, on 40th and 4th day the attendance was same and the attendance were 790 and 214.
Learn more about the substitution method here:
https://brainly.com/question/14619835
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