Please I need help with this question

If you run 10 feet in 3 seconds how long does it take you to run a mile?

mel-nik [20]

It would take you 15 minutes

3 0

3 years ago

A group of artists is selling small prints to raise money for charity. They sell p prints, and make a profit of f(p) on the sale

MrRissso [65]

Answer:

Statement A is true.

Step-by-step explanation:

A. If the group sells 15 prints, they will lose $85.

B. If the group sells 12 prints, they will lose $204.

C. If the group sells 35 prints, they will make $935.

D. If the group sells 28 prints, they will lose $136.

f(p) = 17p - 340

p = prints

f(p) = profit

Let's check the statements to see which one is true.

A. 17.15 - 340 = - 85 OK!

B. 17.12 - 340 = -136 (false, they will lose $136)

C. 17.35 - 340 = 255 (false, they will make $255)

D. 17.28 - 340 = 136 (false, they will make $136)

3 0

3 years ago

Quick I need an answer to this Queston:

fiasKO [112]

Step-by-step explanation:

$- \frac{3}{4} x + 2 \frac{1}{4 } = \frac{1}{2} x + 1$

$- \frac{3}{4} x + \frac{9}{4} = \frac{1}{2} x + \frac{2}{2}$

$\frac{ - 3x + 9}{4} = \frac{2x + 4}{4}$

$- 3x + 9 = 2x + 4$

$- 3x - 2x = 4 - 9$

$- 5x = - 5$

$x = \frac{ - 5}{ - 5}$

$x = 1$

The answer is C.

6 0

2 years ago

What is the range of the function y = x??<br> all real numbers<br> X30<br> Oy0

lara31 [8.8K]

Answer:

2 second one cuz x can't be smaller than 0 because x is SQUARED

3 0

2 years ago

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

so

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)

so

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

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

3 years ago