What is the product?

A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul

N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$ .

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

6 0

4 years ago

BRAINLIESSSTTTT ASAP!!!!!

Yakvenalex [24]

Answer:

D. Corresponding.

Step-by-step explanation:

Interior angles are always on the inside of a set of lines, an example being that if the answer had been interior angles, the top orange line would most likely move just below the straight line it is resting upon and turn so that it would look something like the first picture attached. The picture attached shows a. alternate interior. They can be considered alternate due to the fact that they are on two different sides of that diagonal line.

Exterior angles would be where they are placed in the drawing you provided now, except alternate exterior angles would be on alternating sides of that significant diagonal line. A picture is attached and shows what b. alternate exterior might look like.

Vertical angles appear on opposing sides. This one is hard to word, but I will attach a picture if it allows me to. You see in your picture where the top orange line is resting? Jump across that line. Jump one down in that little acute angle place that is still touching the top horizontal line. This is where you would have had to place your bottom orange angle to make it vertical.

Corresponding angles are just like shown in the question you were given. With corresponding angles, they are always on the same side of the diagonal lines splitting the sides, yet one will seem to rest on the interior and the other will rest on the exterior so that they both look exactly the same curve wise. I would provide a chart online since this is not really one to draw since you already have, but I do not want this answer taken down for plagiarism. Just look up corresponding angles online and it should show you what I mean.