At the movie theatre, child admission is 6.20

1 answer:

3 0

6.20c+9.60a=1016.60
c+a=136

a=136-c
6.20c+9.60(136-c)=1016.60
6.20c+<span>1305.60- 9.60c=1016.60
</span>-3.40c = 1016.60 - 1305.60
-3.40c=-289
c=289/3.4 = 85

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F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$