PLZ HELP QUICK!!! MARK BRAILIEST!!!

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

so

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

so

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago

Pleeease help!!

stepan [7]

The answer to this is y = 7x + 15

8 0

3 years ago

A ladder leans against a building forming an angle of 50° with the ground. The base of the ladder is 8 feet from the building. W

Aleksandr-060686 [28]

Answer:

The correct answer is 12.446 feet.

Step-by-step explanation:

A ladder leans against a building forming an angle of 50° with the ground.

The base of the given ladder is 8 feet away from the building.

Let the length of the ladder that is to be found be x feet.

Thus we find the cosine of the angle of inclination.

∴ cos 50° = $\frac{8}{x}$

⇒ x = 12.446 feet

Therefore the length of the ladder is 12.446 feet.

6 0

3 years ago

Can somebody help me with 17 and 18 please

AveGali [126]

17. (x^2-x+3)(4x+1)
18.(x+2)(x) - (7)(3)

3 0

3 years ago

Solve for x. Thank you.

Bas_tet [7]

Answer:

D) 50/3

Step-by-step explanation:

6/10 = 10/(6+x-6)

6/10 = 10/x

6x = 100

x = 100/6

x = 50/3

4 0

3 years ago