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alexgriva [62]
2 years ago
15

The volume of a water walking ball is $\frac{4}{3}\pi$

Mathematics
1 answer:
saw5 [17]2 years ago
6 0

If the radius of the ball is 1 meter. Then the diameter of the ball will be 2 meters.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

The volume of a water walking ball (in cubic meters) is given below.

\rightarrow\dfrac{4}{3}\pi

Then the diameter of the sphere will be

We know the volume of the sphere is given as

V = \dfrac{4}{3}\pi r^3

where r is the radius of the ball.

Then the radius of the ball will be

\rm \dfrac{4}{3}\pi  = \dfrac{4}{3}\pi r^3\\\\r^3 \  = 1\\\\r \ \ = 1\ m

Then the diameter of the ball will be

d = 2r

d = 2 × 1

d = 2 m

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

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Answer:

equation~ 2x + 5= 2X + 8

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Step-by-step explanation:

equation: 2x + 5 =2x +8

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 4x + 5= 8

   4X= 13

x= 0.3

value of X: 0.3

Perimeter: 13

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If 1 apple cost 3 dollars how much would 5 cost?
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Find the average rate of change of f(x)=4x^2-6 on interval [1,a]​
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Answer:

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Step-by-step explanation:

The average rate of change of f(x) on the interval is ...

  (f(a) -f(1))/(a -1)

  = ((4a² -6) -(4(1)² -6))/(a -1) = (4a² -4)/(a -1)

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The average rate of change on the interval is 4a +4.

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Please help, Brainliest if correct!!!
Lubov Fominskaja [6]

See the attached pic below

In the triangle formed by the statue and the boat, we have

\tan23^\circ=\dfrac x{110+y}

and

\tan38^\circ50'=\dfrac xy

We can solve either equation for y, then substitute that into the other equation. First, let's abbreviate t_1=\tan23^\circ and t_2=\tan38^\circ50'. Then

y=\dfrac x{t_2}

\implies t_1=\dfrac x{110+\frac x{t_2}}

\implies\dfrac{t_1}{t_2}=\dfrac x{110t_2+x}

\implies(110t_2+x)\dfrac{t_1}{t_2}=x

\implies110t_1=\left(1-\dfrac{t_1}{t_2}\right)x

\implies x=\dfrac{110t_1}{1-\frac{t_1}{t_2}}=\dfrac{110t_1t_2}{t_2-t_1}

\implies\boxed{x\approx99}

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4 years ago
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