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Whitepunk [10]
3 years ago
10

David, a platform diver, dives into the pool during practice. The height of David above the water at any given time, s, can be m

odeled by the quadratic function h(s).
Each of the following functions is a different form of the quadratic model for the situation above. Which form would be the most helpful if attempting to determine the time required for David to enter the water?

A. h(s) = -4.9(s - 2)(s + 1)

B. h(s) = -4.9s(s - 1) + 9.8

C. h(s) = -4.9(s - 0.5)2 + 11.025

D. h(s) = -4.9s2 + 4.9s + 9.8

Mathematics
1 answer:
ella [17]3 years ago
4 0

Check the pictures below.

if we knew the roots/solutions of the equation, we can set h(s) = 0 and solve for "s" to find out how many seconds is it when the height is 0.

if you notice in the first picture, when f(x) = 0, is when the parabola hits a root/solution or the ground, for David he'll be hitting the water surface, and the equation that has both of those roots/solutions conspicuous is

h(s) = -4.9(s - 2)(s + 1).

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The 3 numbers are 17,19 and 21

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3 0
3 years ago
I don't know how to do this or what i'm doing plz help
mote1985 [20]

recalling that d = rt, distance = rate * time.


we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?


\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}


so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.


\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}


\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}

7 0
3 years ago
PLEASE HELP! 40 POINTS
Vedmedyk [2.9K]

I'd start with setting up the equations. Calling x the number of students in most popular country 1, and y in country 2.

x + y = 44740

x - y = 5672

now solve for x1 and x2. You can subtract second from first:

0 + 2y = 39068

y = 19534

so

x = 5672 + 19534 = 25206

Add a sentence to answer it in words.

7 0
3 years ago
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Hmm if you tryna find the answer it’s 360 because you are multiplying 15x24 and it equals 360 :)) hope this helps!! If not I’m sorry
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