Mathematics, 03.08.2019 20:30 binodkharal2048

An object is launched from the ground. the object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t2, where t is the time, in seconds, since the object was launched. when will the object hit the ground after it is launched? explain how you found your answer.

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h(t) = 80t – 16t^2
When you are on the ground we have zero height, h(t) = 0. This is saying that at some amount of time we will call t, the object will hit the ground. Now we solve for t to get that time
so we sub h(t) for 0.
0=80t-16t^2 move the -16t^2 to the other side
16t^2=80t divide both sides by t
16t=80 divide both sides by 16
t=5
Now... it is important to realize that there are going to be 2 answers to this equation as you have a t^2. The other answer is 0. Obviously, unless the object started on the ground, the time to hit the ground is not 0 and that answer is ruled out.

Answer from: Disd112984

The object will hit the ground after 5 seconds. You can rewrite the quadratic function as a quadratic equation set equal to zero to find the zeros of the function 0 = –16t2 + 80t + 0. You can factor or use the quadratic formula to get t = 0 and t = 5. Therefore, it is on the ground at t = 0 (time of launch) and then hits the ground at t = 5 seconds.

Answer from: Quest

518,400

step-by-step explanation:

the area is multiplying all the given measurements together.

40 * 18 * 30 * 24 = 518,400

Answer from: Quest

yes

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8. 4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

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An object is launched from the ground. the object’s height, in feet, can be described by the quadrat...