Find an equation of the line with the given properties slope=3 containing the print(-2,3)

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

Find the area. Simplify your answer.

Shalnov [3]

9514 1404 393

Answer:

x^2 +8x +12

Step-by-step explanation:

The area is the product of length and width. The product can be simplified using the distributive property.

A = LW

A = (x +6)(x +2) = x(x +2) +6(x +2)

A = x^2 +2x +6x +12

A = x^2 +8x +12

6 0

3 years ago

Read 2 more answers

Which is the same as dividing 23 by 7

Grace [21]

7 dividing by 23 I'm totally sure

6 0

3 years ago

What is the first step in solving this problem:<br> -2 •(3+8+4)2

blondinia [14]

The first step is to solve the parenthesis because of PEMDAS.

hope this helps,, have a nice day :))

8 0

3 years ago

Read 2 more answers

How do I answer this Question? It’s phythagoras theorem!

zhuklara [117]

Answer:

C = 11.18 D= 20

Step-by-step explanation:

Let's start with C

Square root both areas of the squares

$\sqrt{25} =5$

$\sqrt{100}=10$

Then use the Pythagoras Theorem: $a^{2} +b^{2} =c^{2}$

$10^{2} +5^{2} =125$

$\sqrt{125} =11.18033989$ = C

Moving on to D

Square root both areas of the squares

$\sqrt{841} =29$

$\sqrt{441} =21$

29 is the hypotenuse, we are trying to find the other leg, so:

$29^{2} -21^{2} =400$

$\sqrt{400} =20$ = D

8 0

3 years ago