Can someone help me fill this out

A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32

olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

32*t = 32

t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

4 0

3 years ago

Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was

andrezito [222]

Answer:

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 100, p = 0.42$

92% confidence level

So $\alpha = 0.08$ , z is the value of Z that has a pvalue of $1 - \frac{0.08}{2} = 0.96$ , so $Z = 1.75$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.3336$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.5064$

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

4 0

3 years ago

Here r r my q i have left

enyata [817]

Answer:

D

Step-by-step explanation:

-2 is the y intercept of the first system

8 0

3 years ago

Read 2 more answers

Not sure how to solve 2sinxcosx+cosx=0 Could you help?

kaheart [24]

$2sinxcosx+cosx=0\\\\cosx(2sinx+1)=0\iff cosx=0\ \vee\ 2sinx+1=0\\\\cosx=0\ \vee\ 2sinx=-1\ /:2\\\\cosx=0\ \vee\ sinx=-\frac{1}{2}\\\\x=\frac{\pi}{2}+k\pi\ \vee\ x=-\frac{\pi}{6}+2k\pi\ \vee\ x=\frac{7\pi}{6}+2k\pi\ to\ k\in\mathbb{Z}$

6 0

3 years ago

Given: Lines a and b are parallel and line c is a transversal. Prove: Angle2 is supplementary to Angle8 Horizontal and parallel

kondaur [170]

Answer:

The correct option is;

Corresponding angles theorem

Step-by-step explanation:

Type of lines of lines a and b = Horizontal and parallel lines

The transversal to a and b = Line c

The angles between a and c labelled clockwise from the upper left quarter segment = 1, 2, 4 and 3

The angles between b and c labelled clockwise from the upper left segment = 5, 6, 8 and 7

Therefore, we have;

Statement ${}$ Reason

1. a║b, c is a transversal ${}$ Given

2. ∠6 ≅ ∠2 ${}$ Corresponding angles theorem

3. m∠6 = m∠2 ${}$ Definition of congruent

4. ∠6 is supp. to ∠8 ${}$ Definition of linear pair

5. ∠2 is supp. to ∠8 ${}$ Congruent supplement theorem

Corresponding angles are the angles located in spatially similar or matching corners of two lines that have been crossed by the same transversal. When the two lines having a common transversal are parallel, the corresponding angles will be congruent.

4 0

3 years ago