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3 years ago

I'll give 99 points

Mathematics

2 answers:

photoshop1234 [79]3 years ago

8 0

Maximum height is at dH/dt=0

H= -16T^2+80T+5

dH/dt=-32T+80

-32T+80=0

32T=80

T=2.5s when height is max

Hmax=-16(2.5)^2+80*2.5)+5

=105ft

catching the ball at 5' means solve for T at H=5

H= -16T^2+80T+5=5

-16T^2+80T=0

-16T(T-5)=0

so T=0 cannot be the ans or T=5s

Leto [7]3 years ago

3 0

<span>H= -16T^2+80T+5
so its vertex is at T=b/2a=80/(2*(-16))=2.5
H(2.5)=-16*2.5^2+80*2.5+5=105
last one
H=5
</span>-16T^2+80T+5=5
-16T^2+80T=0
T=0 when ball is thrown or T=5 when it is caught

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densk [106]

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3 years ago

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algol [13]

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3 years ago

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

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

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zhuklara [117]

Answer: B) y=6/11.

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Using L’Hôpital’s rule:
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