Luis deposits an amount of $2000 into an account. The account returns 2.4% interest, compounded quarterly.

Find the volume of the cylinder. Express your answer in terms of pi. The cylinder has a radius of 8 inches and a height of 10 in

gulaghasi [49]

Answer:

640π

Step-by-step explanation:

Multiply 10 by 64 [8²], then attach π to it.

6 0

3 years ago

A measuring cylinder of radius 3cm contains water to a height of 49cm. if this water is poured into a similar cylinder of radius

levacccp [35]

Hey there! I'm happy to help!

First, let's find the area of the circle of the first cylinder. To do this, we square the radius and multiply by 3.14.

3²=9

9×3.14=28.26

Now we multiply by our height of 49 cm to see how much water there is.

28.26×49=1384.74

Now, let's find the area of the circle of the 7 cm cylinder.

7²=49

49×3.14=153.86

Now, we will take our liquid amount (volume) and divide by this circle amount (base) to find the number of centimeters of height of the water column.

1384.74÷153.86=9

So, the height of our water column will be 9 cm.

Have a wonderful day and keep on learning! :D

5 0

3 years ago

Order from least to greatest 0, 12, 1/8, 0.135, 13%

klio [65]

Answer:

0.135, 1/8, 0, 13%, 12

6 0

3 years ago

Read 2 more answers

1

Mariulka [41]

Answer:

27.0 feet

Step-by-step explanation:

Length of an arc (s) = 2πrθ/360................. Equation 1

Where r = radius of the circle, θ = angle in degree substends by the arc at the center of the circle.

From the question,

Given: r = 9 feet, θ = 3 radian = 3 (57.2950) = 171.89°

Constant: π = 3.14

Substitute these values into equation 1

s = 2(3.14)(9)(171.89)/360

s = 9715.2228/360

s = 26.99 feet

s ≈ 27.0 feet

7 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

$0.028x=0.68$

$x=24.3$

Alternative method

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