Help I will be marking brainliest!!!

A conical paper cup has a radius of 2 inches. Approximate, to the nearest degree, the angle β so that the cone will have a volum

dezoksy [38]

Volume = area of base *height

20 = pir^2 *h

20 = 3,14*4*h

h = 20/12,56 = 1,59 rounded 1,6

- using the triangle formed by radius,the height of cup and the slant height of cup so we can writing that

tan ß = h/r so tan ß = 1,6/2 = 0,8

tan ß = 0,8 so ß = arctan 0,8 = 38,6 degrees so 39 degrees

hope helped

5 0

3 years ago

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

In Mrs. Trach’s class, 18 students prefer to do their homework before dinner. In Ms. Love-Benson's class, 20 students prefer to

deff fn [24]

Yes because in Mrs. Benson‘s class there are 20 students and in Mrs Tracy’s class there are only 18

5 0

3 years ago

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OK I need help the answer is c but I didn't understand the teacher explanation pls help me

serg [7]

9 - 6 + 4 - 8/3 ..,
geometric series a(n) = a1r^(n-1)

r = a(n+1)/a(n)
-6/9 = -2/3
4/-6 = -2/3
-8/3/4 = -2/3
so r = -2/3 and a1 = 9

Sn = a1(1-r^n)/(1-r) = 9(1-(-2/3)^n)/(1-(-2/3))
n is infinite Sn = 9/(5/3) = 27/5

7 0

3 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

ch4aika [34]

Answer:

First answer.

Step-by-step explanation:

We know that a complete rotation would be 360°. If the figure is being rotated 90°, this is a quarter of a complete rotation.

When looking at the given answers, only the first answer seems logical.

5 0

3 years ago

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