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3 years ago

How is the relationship between the formula for the volume of a cone and the formula for the volume of acylinder

1 answer:

4 0

The volume of the cylinder is calculated by using the product of the area of its base by its height. ... Therefore, the volume of a cone is . So, we can take a logical conclusion: “the volume of a cone means the third part of the volume of a cylinder having the same base and the same height”.

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I need the answer to this fast Please

jeyben [28]

The initial function is slope

=> slope = 7-1/6-0 = 6/6 = 1

Hope this helps you :)

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2 years ago

Write 6 : 9 :: 2 : 3 as a fraction proportion.

Firdavs [7]

6/9 (2/3) and 2/3. Hope this helps you.

3 0

3 years ago

Write 3 addition facts that have the same sum as 5 + 9

Dovator [93]

Answer:

10+4=14

8+6=14

13+1=14

5 0

3 years ago

Read 2 more answers

A 4-ft-high and 7-ft-wide rectangular plate is submerged vertically in water so that the top is 1 ft below the surface. Express

Readme [11.4K]

Total depth of the bottom of the plate is 4 + 1 = 5

Force = limit(5,1) 62.5 *7* x * dx

= 437.5. Lim(5,1) x*dx

= 437.5(x^2/2)^5 , 1

= 437.5 x (5^2/2 - 1/2)

= 437.5 x 12

= 5,250 pounds

6 0

4 years ago

Yea I can do tomorrow it would take me to get the money back to you

Tems11 [23]

Given, the equation that represents the height of an object:

$y(t)=100t-16t^2$

First, we will find the velocity of the object which is the first derivative of the height using the method of the limits

$\frac{dy}{dt}=\lim_{h\to a}\frac{y(3+h)-y(3)}{(3+h)-(3)}$

We will find the value of the function y(t) when t = 3, and when t = 3+h

$\begin{gathered} y(3+h)=100(3+h)-16(3+h)^2 \\ y(3+h)=300+300h-16(9+6h+h^2) \\ y(3+h)=300+300h-144-96h-16h^2 \\ y(3+h)=156+4h-16h^2 \\ y(3)=100(3)-16(3)^2=156 \end{gathered}$

Substitute y(3+h) and y(3) into the expression of the limit

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}\frac{156+4h-16h^2-156}{3+h-3}=\lim_{h\to a}\frac{4h-16h^2}{h} \\ \\ \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \end{gathered}$

Where a = 0

d) compute the instantaneous velocity at t = 3

$\frac{dy}{dt}|_{t=3}=100-16*2*3=4$

So, the answer will be:

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \\ a=0 \\ \\ \frac{dy}{dt}|_{t=3}=4\text{ ft/sec} \end{gathered}$

5 0

1 year ago