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2 years ago

The volume of a spherical ball is 4500rr cubic centimeters. Find the radius of the ball.

Mathematics

1 answer:

andre [41]2 years ago

4 0

Answer:

its is6700rr centimeters

Step-by-step explanation:

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Carlotta has 12 2/3 yd of ribbon to be divided among 19 girls in the cheerleading squad.how much ribbon will each girl get if it

trasher [3.6K]

Answer:

Each girl gets 2/3 yd of ribbon.

Step-by-step explanation:

12 2/3=38/3

(38/3)/19

(38/3)(1/19)

(2/3)(1/1)=2/3

4 0

3 years ago

Eric wants to find out the most popular after-school activity among high school seniors at his school. What is the population in

damaskus [11]

Answer: The population in Eric's survey is the high school seniors.

3 0

3 years ago

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The length of a rectangular field is 24 meters this is 3 meters less than twice the width find the width

stepladder [879]

Answer:

The width is 5 meters

Step-by-step explanation:

The rectangular field is 24 meaning adding just one width and length of the rectangle would be 12.

x = width

2x - 3 + x = 12

3x - 3 = 12

3x - 3 + 3 = 12 + 3

3x = 15

x = 5

5 0

2 years ago

The Venn diagram shows all of the elements in sets A, B and ξ. Find the elements in (A ∩ B) U (A U B)’

marusya05 [52]

This is impossible to answer without knowing what the sets A and B contain (and what ξ even refers to - universal set?).

However, we have

(A U B)' = A' ∩ B'

so that

(A ∩ B) U (A U B)' = (A ∩ B) U (A' ∩ B') = (A U A') ∩ (B U B')

If ξ is indeed the universal set, then both A U A' = ξ and B U B' = ξ, so we end up with ξ ∩ ξ = ξ.

8 0

2 years ago

A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its op

Alina [70]

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, $\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h$

Volume of cone is given by

$V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3$

By differentiating with respect to time t, we get that

$\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}$

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie $\dfrac{dv}{dt}=-4\ in^3$

and h = 2 inches deep.

$-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}$

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

7 0

2 years ago