Help!!!!!!!! Plz I need help

The image shows three tennis balls enclosed in a cylindrical can. Choose all that are correct. The volume of a single tennis bal

Fynjy0 [20]

Answer:

The volume of a single tennis ball is $14.14\ in^{3}$

The volume of three tennis balls is $42.42\ in^{3}$

Step-by-step explanation:

Step 1

Find the volume of a single tennis ball

we know that

The volume of a sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

where r is the radius of the sphere

In this problem

$r=3/2=1.5\ in$

substitute

$V=\frac{4}{3}\pi (1.5)^{3}$

$V=14.14\ in^{3}$

Step 2

Find the volume of the can

we know that

the volume of the cylinder is equal to

$V=\pi r^{2} h$

where r is the radius of the cylinder

h is the height of the cylinder

in this problem we have

$r=3/2=1.5\ in$

$h=8.4\ in$

substitute

$V=\pi (1.5)^{2}(8.4)$

$V=59.38\ in^{3}$

Statement

case A) The volume of a single tennis ball is $14.14\ in^{3}$

The statement is true

See the procedure in Step 1

case B) The volume of the can is $56.55\ in^{3}$

The statement is false

The volume of the can is $59.38\ in^{3}$ --> see the procedure Step 2

case C) The empty space inside the can is $14.13\ in^{3}$

The statement is false

To find the empty space subtract the volume of three tennis ball from the volume of the can

$59.38\ in^{3}-3*14.14\ in^{3}=16.96\ in^{3}$

case D) The volume of three tennis balls is $42.42\ in^{3}$

The statement is true

To find the volume of three tennis balls multiply the volume of a single tennis ball by three

$14.14*3=42.42\ in^{3}$

8 0

3 years ago

At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

$f(x) = x^3-2x^2-4x+1$ in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

$f(1) = 1-2-4+1 = -4\\f(0) = 1$

The value

$\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5$

Find derivative for f

$f'(x) = 3x^2-4x-4$

Equate this to -5 to check mean value theorem

$3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1$

We find that 1/3 lies inside the interval (0,1)

4 0

3 years ago

In a lottery the top cash prize and $663 million going to three lucky winners players pick five different numbers from 1 to 58 a

user100 [1]

Answer:

43

Step-by-step explanation:

5 0

3 years ago

HI can you help me with this problem?The quantity demanded x for a certain brand of MP3 players is 300 units when the unit price

Vlad [161]

The demand equation is P = -0.05x + 105

STEP - BY - STEP EXPLANATION

What to find?

The demand equation.

The following information is used to solve the problem.

If (x₁ , y₁) and (x₂, y₂) are any dinstinct points on a line a non-vertical line L, then the equation of the line that has a slope(m) and passes through the point (x₁ , y₁) is given by the formula below:

$y-y_1=_{}\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Assume p, be the unit price and x be the demand when the price is $p.

From the question given, when the unit price x = 300 units, the demand p = $90 and when the unit price x = 1300 units, the demand p = $40

This means the demand curve passes through the points (300, 90) and (1300, 40).

Assume the demand equation is linear, then the line of the equation passes through the points (300, 90) and (1300, 40).

Now;

y=p x₁ =300 y₁ = 90 x₂ = 1300 y₂=40

Substitute the values into the function and simplify.

$p-90=\frac{40-90}{1300-300}(x-300)$

Simplify the above.

$p-90=\frac{-50}{1000}(x-300)$ $p-90=-0.05(x-300)$

Open the parenthesis.

P- 90 = -0.05x + 15

Add 90 to both-side of the equation.

P = -0.05x + 15 + 90

P = -0.05x + 105

Hence, the demand function is P = -0.05x + 105

6 0

1 year ago

Jerry pays $1.40 per gallon for premium gasoline. He spends $16.80 and wants

muminat

Answer:

let x be the total amount of gallons he bought

$16.80÷$1.40 = x

16.80÷1.40=12

x=12

Step-by-step explanation:

3 0

3 years ago