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Mathematics 24 Online
OpenStudy (anonymous):

I really need help with a worded answer quiz please help me I need to finish this so I can move on with my other school work (:

OpenStudy (anonymous):

mhmm

OpenStudy (anonymous):

@apollard99 if u dont get it its fine i dont either

OpenStudy (anonymous):

Sorry, im not good with this.

OpenStudy (anonymous):

@potato_donuts

OpenStudy (anonymous):

the ball in the picture looks like it goes up to a height of about 20 or 21 feet, hard to see exactly

OpenStudy (anonymous):

the ball given by the height \[ h=–16t^2 + 48t + 15 \] has a maximum at the second coordinate of the vertex the first coordinate of the vertex is \[-\frac{b}{2a}=\frac{48}{32}=\frac{3}{2}\]

OpenStudy (anonymous):

if you put in \(\frac{3}{2}\) for \(t\) you get \(51\) which is larger than 21

OpenStudy (anonymous):

i will leave the "complete sentences" up to you, but the answer is what i wrote above

OpenStudy (anonymous):

okay thank you! yeah im very good with writing just not the math part :P can u help me with parts 2 and 3?

OpenStudy (anonymous):

what are they?

OpenStudy (anonymous):

the equation \(h=-16t^2+48t+15\) means at time \(t=0\) you have \(h=15\) i.e. you are 15 feet up seems like a platform to me in the graph when \(t=0\) you are about at 5 feet up so standing on the floor

OpenStudy (anonymous):

as for the height of the platform, since the person not standing on the platform throws the ball from a height of 5, and the person on the platform throws from a height of 15 , i would estimate the height of the platform at \(15-5=10\) feet up

OpenStudy (anonymous):

site is really slow, hope you got that

OpenStudy (anonymous):

yeah im noticing that it took me like 3 minutes to be able to fan you but yes im getting all of this

OpenStudy (anonymous):

it is increasing where the curve is going up

OpenStudy (anonymous):

in the graph that is from \(t=0\) to \(t=1\)

OpenStudy (anonymous):

in the equation from \(t=0\) to \(t=\frac{3}{2}=1.5\)

OpenStudy (anonymous):

Thank you (:

OpenStudy (anonymous):

as for the rate of change the ball in the graph goes from 5 to 20 one second, so the rate of change is 15 feet per second on average

OpenStudy (anonymous):

the one in the equation goes from 15 to 51 in 1.5 seconds, so the rate of change is \(51\div 1.5= 34\) feet per second

OpenStudy (anonymous):

oops mistake!!

OpenStudy (anonymous):

rate of change is \((51-15)\div 1.5=36\div 1.5=24\) feet per second

OpenStudy (anonymous):

gotta run, good luck

OpenStudy (anonymous):

thanks for this (:

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