Ask question

Ask question

All categories

3 years ago

6

A ball is thrown strait up from the top of a 64 foot tall building with an initial speed of 48 feet per second. The height of th

e ball as a function of time can be modeled by the function h(t)=-16t^2+48t+64. At what other time will the ball be at a height of 64 feet? How long will it take for f ball to reach its maximum height? What is he maximum height reached by the ball? How long willlit take for the ball to hit the ground?

1 answer:

Mrrafil [7]3 years ago

4 0

Answer:

see below

Step-by-step explanation:

h(t)=-16t^2+48t+64

Find when h(t) = 64

64 = -16t^2+48t+64

Subtract 64 from each side

0 = -16t^2+48t

Factor

0 = -16t( t-3)

Using the zero product property

-16t =0 t-3 =0

t=0 and t=3

Other than at t=0, so at t=3 seconds

How long will it take for f ball to reach its maximum height?

Find the vertex

t = -b/2a = -48 / ( 2 * -16) = -48/ -32 = 1.5 seconds

It will reach the maximum height at 1.5 seconds

What is the maximum height reached by the ball?

The height will be

h(1.5) = -16*(1.5)^2+48(1.5)+64

= -36+72+64

= 100

The maximum height is 100 ft

How long will it take for the ball to hit the ground?

h(t) =0

0=-16t^2+48t+64

Factor

0 = -16( t^2 - 3t -4)

0=- 16( t-4) (t+1)

Using the zero product property

t-4=0 t+1=0

t =4 t = -1

Since time cannot be negative

t=4 seconds

You might be interested in

From measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert the area into squa

NemiM [27]

So the problem ask to find and convert the area into square meters. So to convert is you must do the cross multiplication process that could cancel out unit and made the answer into a square meter, so the answer would be 1. x10^-12m^2. I hope you are satisfied with my answer and feel free to ask for more

4 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

A and B are complementary angles. If A = (x – 23)° and m

Lana71 [14]

Answer:

∠ B = 85°

Step-by-step explanation:

Complementary angles sum to 90° , then

x - 23 + 2x + 29 = 90 , that is

3x + 6 = 90 ( subtract 6 from both sides )

3x = 84 ( divide both sides by 3 )

x = 28

Then

∠ B = 2x + 29 = 2(28) + 29 = 56 + 29 = 85°

5 0

3 years ago

Repost of my question i need help quicklyy help plzz

GuDViN [60]

Answer:

According to logarithmic properties.... The right hand side can be written as

log base 7 (180/3).....which is log base 7 60

So according to the question cancel out the log base 7 from both sides....

Then we get 8r + 20 = 60

That is 8r = 40..

That is r = 5......

Therefore the value of r is 5

8 0

3 years ago

I need an answer quick!!!!

Illusion [34]

Answer:

investment 2 $400 Simple Interest Rate

Step-by-step explanation:

5 0

3 years ago