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Alexeev081 [22]
3 years ago
5

kayla is standing in the gym watching the PE teacher toss volleyballs into the ball bin. She estimates that the PE teacher's han

d is 4 feet above the gym floor, the ball bin is 4 feet high and she is standing 10 feet from the ball bin while the maximum height of the ball is 8 feet. quadratic equation.
Mathematics
1 answer:
miss Akunina [59]3 years ago
3 0

Answer:

pp

Step-by-step explanation:

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A shopping container has a volume of 2,800 cubic inches. which could be the dimensions of the container? WILL GET BRAINLIEST​
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Missing part of the question

- 10 in, by 12 in. by 24 in.

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The possible dimensions of the container are

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I hope i helped you. have a wonderful day!

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F(x)=3x+1 and g(x)x^2 what would gf(x) be
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Jonah took a survey. Of the students he surveyed, 40% of the students play baseball. Of those students, 60% also play soccer. If
Flura [38]
Let s represent the number of students surveyed.
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8 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
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