−2/3p+1/5−1+5/6p, Combine Like Terms.

A cylinder and a cone have the same radius and height. The volume of the cylinder is 639 ft ^3. what is the volume of the cone?

AURORKA [14]

The volume of the cone is: A. 213 ft³.

<h3>Volume of a Cylinder and a Cone</h3>

Volume of cylinder = πr²h.
Volume of cone = ⅓πr²h.

Given:

radius and height of both the cylinder and the cone are equal.

Volume of Cylinder = 639 ft³

Therefore:

πr²h = ⅓πr²h

This means that volume of cone would be ⅓ of the volume of the cylinder.

Volume of cone = ⅓(639) = 213 ft³

Therefore, the volume of the cone is: A. 213 ft³.

Learn more about volume of cone on:

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7 0

2 years ago

On January 31, Jean Marie’s business receives a bill for that month’s utilities in the amount of $500. Jean sets it aside becaus

marissa [1.9K]

On January 31, Jean Marie’s business receives a bill for that month’s utilities in the amount of $500. Jean sets it aside because she does not plan to pay the bill until its due date of February 15. What effect, if any, does this event have on the company’s accounting equation as of January 31?

Solution: The event that Jean does not plan to pay the bill until due date of February 15 must be recorded. Recording this event would increase the liabilities and decrease equity on January 31.

4 0

2 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago

State whether the lines are parallel, perpendicular, coinciding, or neither: 6x+3y=–15 and y–3=–2x

lianna [129]

Answer:

Parallel

<u>Step-By-Step Explanation:</u>

Put the Function in Slope Intercept Form and Find the Slope of 6x+3y = 15

6x+3y = 15

3y = -6x + 15

3y/3 = -6x/3 + 15/3

y = -2x + 5

<u>We can see that the slope of 6x+3y = 15 is -2</u>

Put the Function in Slope Intercept Form and Find the Slope of y–3=–2x

y–3=–2x

y = -2x + 3

Here are our two Functions In Slope Intercept Form

y = -2x + 5

y = -2x + 3

<u>Remember the m = slope and the b = y-intercept</u>

y = mx + b

y = -2x + 5

y = -2x + 3

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We can see both equations have the same slope of -2 so this means they could be parallel because parallel functions have the same slope but coinciding functions have the same slope too. To tell if the two functions are coinciding, the functions need to have the same slope and the same y-intercept. Looking at the two functions, we can see they have the same slope of -2 but their y-intercept are different so this makes the two functions parallel.

4 0

3 years ago

Jim had a box full of baseball cards on Monday he gave 1/3 of them to John on Tuesday he gave 10 of them to Jeff finally on Wedn

larisa [96]

Answer:

120 baseball cards

Step-by-step explanation:

15 × 2 = 30

30 + 10 = 40

40 ÷ 1/3 = 40 × 3/1 = 120

Therefore there were 120 cards in the back

<em>I just did the inverse of each process starting from the last step and ending at the first</em>

5 0

3 years ago

−2/3​p+1/5​−1+5/6​p, Combine Like Terms.

−2/3p+1/5−1+5/6p, Combine Like Terms.