Mathematics, 25.07.2019 01:30 jaceechesick

Aball is projected in to the air. its height at time t is given by the equation h = -16t^2 + 60t + 1. solve for the time as which the height will be 8 feet. show your work and explain the steps you used to solve. round your answer to the nearest hundredth.

Answers: 1

Show answers

Answers

Answer from: getsic

The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:

$h=-16 t^{2} +60t+1$
$8=-16 t^{2} +60t+1$
$16 t^{2} -60t-1+8=0$
$16 t^{2} -60t+7=0$

This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:
$t= \frac{-bplus minus \sqrt{ b^{2}-4ac } }{2a}$
You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."

In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of) $t^{2}$ = 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7

Substituting these into the quadratic formula gives us:
$t= \frac{-(-60)plus minus \sqrt{ (-60)^{2}-4(16)(7) } }{2(16)}$
$t= \frac{60plus minus \sqrt{3600-448 } }{32}$
$t= \frac{60plus minus \sqrt{3152 } }{32}$

As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:
$t= \frac{60+56.1426}{32} =3.63$
and
$t= \frac{60-56.1426}{32} =.12$

So the height is 8 feet at t = 3.63 and t=.12

It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.

Answer from: Quest

21.00 per mile and plus 5.50 is 26.50 i think that im pretty sure it is right

Answer from: Quest

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: