Find each missing measure.

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5<img src="https://tex.z-dn.net/?f=t%5E%7

DanielleElmas [232]

Answer:

Part A)

No

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation $\displaystyle h(t)=-5t^2+14t+2$ models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

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

Our quadratic is:

$\displaystyle h(t)=-5t^2+14t+2$

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}$

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

$\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2$

Evaluate:

$\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}$

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

$0=-5t^2+14t+2$

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Again, a=-5, b=14, and c=2. Substitute appropriately:

$\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}$

Evaluate:

$\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}$

We can factor the square root:

$\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}$

Hence:

$\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}$

Divide everything by -2:

$\displaystyle x=\frac{7\pm\sqrt{59}}{5}$

Hence, our two solutions are:

$\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362$

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

5 0

2 years ago

Read 2 more answers

6/13=3/c what I the answer and show me how you got your answer.show me how you did it.

umka21 [38]

6 3
---- * ----
13 c

this is a proportion...so we cross multiply
(6)(c) = (3)(13)
6c = 39...divide both sides by 6
c = 39/6 which reduces to 13/2 or 6 1/2 <== ur answer

4 0

2 years ago

Identify the values a, b, and c is the first step in using the quadratic formula to find the solution(s) to a quadratic equation

zlopas [31]

The values are a = 7, b = -9, c = -18.

<u>Step-by-step explanation:</u>

The given quadratic equation is $7x^{2} - 9x = 18$