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3 years ago

What is the domain of the function f(w) = 0.5(35 – 2w)? Explain your answer.

1 answer:

5 0

The domain is all real numbers, the function is a variant of linear, which could be expressed in the form of y=mx+b as y = -w + 17.5, therefore all real numbers is the domain and the range.

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11 less than 7 times a number<br> is 5 more than 6 times the<br> number. Find the number

vivado [14]

Answer:

5 i think. im not good at math. sorry if wrong

4 0

3 years ago

Read 2 more answers

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is $49\ ft$

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

$h(t)=-16t^{2} +63t+4$

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

$h(3)=-16(3)^{2} +63(3)+4=49\ ft$

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

Convert the function into a vertex form

$h(t)=-16t^{2} +63t+4$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$h(t)-4=-16t^{2} +63t$

Factor the leading coefficient

$h(t)-4=-16(t^{2} -(63/16)t)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})$

$h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})$

Rewrite as perfect squares

$h(t)-(67,600/1,024)=-16(t-(63/32))^{2}$

$h(t)=-16(t-(63/32))^{2}+(67,600/1,024)$

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

$h(t)=-16t^{2} +63t+4$

For h(t)=0

$0=-16t^{2} +63t+4$

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0

2 years ago

the area of a rectangular photograph is 35 square inches. If the width of the photo is 5 inches how tall is the photo

Vika [28.1K]

We know that:

Area= length*width

Plug in the stuff you know.

35= length(tall)*5

Divide both sides by 5.

7= length

The photo is 7 inches tall.

I hope this helps!
~kaikers

5 0

2 years ago

Select the correct answer for each statement. (k+2, k+3 , k+4) or (k+10, k+12, k+14) will yield consecutive odd integers. (k+6,

almond37 [142]

Answer:

Step-by-step explanation:

one

The only really certain way of getting an odd integer is to do 2k + 1. Therefore the answer should be something like 2k+1, 2k+3, 2k + 5

The best I can do for the first one is to assume that k is odd and that the answer is k + 10, k+12, k+14, but I would ask your instructor if this is merely the best answer out of a poor lot or if the question really has no answer.

Two things should be noted.

1. k has to be odd.

2. The first choice has to give consecutive integers because what is added on is 2 3 and 4. Those numbers are consecutive.

Two

k+1, k+2, k+3. See point 2 above. k is a constant and any number. 1,2,3 are consecutive so the results are consecutive. Even if k < 0 the numbers will be consecutive.

k= - 10

k+1 = - 9

k+2 = -8

k+3 = - 7

These are consecutive.

Your first choice will give either consecutive odd or even numbers depending on what k is.

If k = even, then k+6, k+ 8, k+10 will all be even.

If k = odd then the givens will be odd.

4 0

3 years ago

Read 2 more answers

Help me plzz plzzz plese2

Dominik [7]

Answer:

122m^2

Step-by-step explanation:

Awnser is in picture along with steps

7 0

2 years ago