2 1/4 case of crayons . Each case holds 24 boxes of crayons . Each box holds 8 crayons . How many crayons

A rocket is launched from a tower what time will the rocket reach its max

Lena [83]

Answer:

Step-by-step explanation:

A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by the function: h(t) = -16t2 + 144t + 160, where ‘t’ represents the time the ball is in the air in seconds and h(t) represents the height, in feet, of the ball above the ground at time t.

a) Graph the function h(t) = -16t2 + 144t + 160 (see below)

b) What is the height of the building?

The height of the building is also the height of the tennis ball before it is launched into the air. This occurs when t=0 so substitute 0 for t and you get:

H(0) = -16(0)2 + 144(0) + 160

The height of the building is 160 feet.

c) At what time did the ball hit the ground?

The ball hits the ground when the height is 0. Therefore, we are looking for a solution to: -16t2 + 144t + 160 = 0

Use the quadratic formula or put this into a calculator. The solution is t=10 and -1, but only 10 makes sense. Therefore, the ball hits the ground at 10 seconds.

d) At what time did the ball reach its maximum height?

You can put this into the calculator or you can realize that the maximum height is also

− the vertex. The x-value (‘t’ in this case) is 2

−144

which is (2)(−16) = 4.5.

Therefore, the ball reached its maximum height at 4.5 seconds.

e) What is the maximum height of the ball?

We calculated the time of the maximum height (4.5 seconds). Therefore, substitute 4.5 into the function to find the maximum height.

-16(4.5)2 + 144(4.5) + 160

The maximum height is 484 feet.

5 0

2 years ago

the line containing the altitude to the hypotenuse of a right triangle whose vertices are p(-1, 1), q(3, 5), and r(5, -5).

finlep [7]

The equation of the straight line is given by 5y-x=6

An equation can be used to describe a straight line drawn on the Cartesian Plane. These equations have a generic structure and can change based on the slope and where the line intersects the axes.

We will greatly simplify this issue by utilizing the fact that this triangle is specified as a right triangle. We would have had to establish the presence of a right angle if it had been presented as a general triangle.

The segment QR is the hypotenuse of the triangle.

Any side's altitude is located along a line that is perpendicular to that side. We are aware that the slopes of perpendicular lines are negative reciprocals, or:

Slope of line joining Q(3, 5) and R(5, -5)

m= $\frac{y_2-y_1}{x_2-x_1}$

Hence slope is= -10/2= -5

Now we know that for two perpendicular line the slope of one line is the negative reciprocal of the other.

Slope of line perpendicular to QR=1/5

This line passes through P(-1,1)

Equation of the straight line that passes through (1,-1) and with a slope of 1/5 is given by:

$(y-y_1)=m(x-x_1)\\$

Substituting the values we get:

y-1=1/5(x+1)

or,5y-5=x+1

or,5y-x=6

Hence the equation of the required line is : 5y-x=6

To learn more about straight lines visit:

brainly.com/question/17757770

#SPJ4

8 0

2 years ago

PLEASE HELP ME ASAP!!!

Olenka [21]

Answer:

The answer is D. Surface Tension. :) I am forsure this is correct but please let me know. If I am correct please rate me 5 and thank me and mark me brainlest :)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Helpppppp! Choose the correct equation: A B C D

Reil [10]

Answer:

C

Step-by-step explanation:

tricky question but i hoped this helped

5 0

2 years ago