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maria [59]
3 years ago
12

A sports ball has a diameter of 12 cm. Find the volume of the ball.

Mathematics
2 answers:
velikii [3]3 years ago
8 0

The volume of a sphere is \frac{4}{3}πr³

If the diameter is 12 cm, divide that in half to get the radius, which is 6.

Now, just plug your variables into the equation for volume.

V = \frac{4}{3}(3.14)6³

V= \frac{4}{3}(678.24)

V= 904.32 cm³

stealth61 [152]3 years ago
5 0
Step 1: I drew a visual representation

Step 2: write the formulae for the volume of a sphere which is V=4*PI*r cubed all divided by 3. Refer to my image

Step 3: answer is 7234.56 cm cubed

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F as a function of x is equal to the square root of quantity 6 x plus 9 , f as a function of x is equal to the square root of qu
Elis [28]

Answer:

A. \sqrt{6x+9} +\sqrt{6x-9}

Step-by-step explanation:

We are given the functions f(x) and g(x) as,

f(x)=\sqrt{6x+9}

g(x)=\sqrt{6x-9}

It is required to find the function ( f+g ) i.e. ( f+g ) (x)

So, (f+g)(x)=f(x)+g(x)

i.e. (f+g)(x)=\sqrt{6x+9}+\sqrt{6x-9}

Hence, we get that (f+g)(x)=\sqrt{6x+9}+\sqrt{6x-9}.

So, the first option is correct.

4 0
3 years ago
Help pleaseeee <br><br> 3.3 - 5.7n – 3n +8= ??
nasty-shy [4]

Answer:

11.3 - 8.7n

Step-by-step explanation:

Its almost basically combining like terms

You subtract - 5.7n - 3n to get -8.7n, then you add 3.3 + 8 to get 11.3 and you finally put it an equation 11.3 - 8.7n

5 0
3 years ago
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Alchen [17]

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Step-by-step explanation:

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6 0
3 years ago
Find the volume of the solid generated when R​ (shaded region) is revolved about the given line. x=6−3sec y​, x=6​, y= π 3​, and
Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

where:

r=6-(6-3 sec(y))

r=3 sec(y)

a=0

b=\frac{\pi}{3}

so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

6 0
3 years ago
Can someone help me with this?
Mandarinka [93]

Answer:

Step-by-step explanation:

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Perimeter of rectangle = 2*(length + width)

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                                   = 2*6x + 2 *4

                                     = 12x + 8

4 0
3 years ago
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