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3 years ago

Should i write a song and post it?

Mathematics

2 answers:

murzikaleks [220]3 years ago

8 0

Answer:

If you want, thats cool

Step-by-step explanation:

Im kinda curious about the song....

thats a nice day

tatuchka [14]3 years ago

7 0

Answer:

yesssssssssssssss do itttttttt

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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

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

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

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

$0 \leq x \leq 48.6\ ft$

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

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

2 years ago

WERE ARE ALL MY GAY QUEENS AND KINGS OUT THERE♥️(love y'all)<br>

Elina [12.6K]

RIGHT HEREEEEEEEEEEEEE

4 0

3 years ago

Read 2 more answers

Stronomers estimate the internal temperature of the sun is 15,000,000  K. What is 15,000,000 written in scientific notation? 1.

svetoff [14.1K]

Answer:

Scientific notation for 15,000,000

$= 1.5 \times {10}^{7}$

6 0

3 years ago

What is a wvgsa test? and how can I study

Dvinal [7]

Answer:

http://wv.portal.airast.org/training-tests.stml

Step-by-step explanation:

copy this link into your url search that should help.

4 0

3 years ago

2) An outdoor staircase has a total vertical elevation of 11 feet from the road level to the front entrance of a building. If th

LekaFEV [45]

2a

tan37=r/8

r=8tan37

r≈6 in rise per step (to nearest whole inch)

2b

11*12/6=22 steps needed to climb the 11 ft to the entrance

...

3a

tanα=14/20

α=arctan(7/10)

α≈35° (to nearest whole degree)

..

3b

tanα=21/17

α=arctan(21/17)

α≈51° (to nearest degree)

3 0

3 years ago