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Vesnalui [34]
3 years ago
11

What is the volume of a cone that has a radius of 4 cm and a height of 9 cm?

Mathematics
2 answers:
vodka [1.7K]3 years ago
8 0
For any sort of pyramid or cone, the volume is 1/3 of the volume of a prism with the same base and height. Since the volume of a prism/cylinder is V=Bh, the volume of a pyramid/cone is V=\frac{1}3Bh.

In this case, our base is a circle, which has a radius of 4 cm.
The area of a circle is A=\pi r^2 where r is the radius.
A=\pi (4)^2=\pi (4\times4)=16\pi=B

We now know that our base is 16π cm.
We also know that our height is 9 cm.
Let's plug these into our volume formula.

V=\frac{1}3\times16\pi \times9

Use 3.14 to approximate pi as the question states. 16 × 3.14 = 50.24.

V\approx\frac{1}3\times50.24\times9

We could punch all of that into our calculator to get the same answer, but since 1/3 of 9 is clearly 3, let's just go with that.

V\approx3\times50.24

\boxed{V\approx150.72\ cm^3}
brilliants [131]3 years ago
5 0
150.8cm³ Hope this helps!!!! :)<span>
</span>
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Given that 18f = 33g, identify the ratio off to g in simplest form.
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11/6

Step-by-step explanation:

18f = 33g

f/g = 33/18 = 11/6

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Answer:

y=3x+5

Step-by-step explanation:

I think that this is what they mean...

y=mx+b

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3 years ago
Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function f(
alekssr [168]

Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function ()=−(+)^−

<em><u>Answer:</u></em>

vertex = (-4, -5)

Axis of symmetry = -4

use the (-4, -5) to find the minimum value

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

<em><u>Solution:</u></em>

Given function is:

f(x) = (x+4)^2 - 5

The equation in vertex form is given as:

y = a(x-h)^2+k

Where, (h, k) is constant

On comparing give function with vertex form,

h = -4

k = -5

Vertex is (-4 , -5)

Axis of symmetry : x co-ordinate of vertex

Thus, axis of symmetry = -4

The coefficient of x^2 is positive in given function.

Thus the vertex point will be a minimum

Minimum\ value = f(\frac{-b}{a})

f(x) = x^2 + 8x + 16 - 5\\\\f(x) = x^2 + 8x + 11

f(x) = ax^2+bx+c

On comparing,

a = 1

b = 8

x = \frac{-b}{2a} = \frac{-8}{2 \times 1} = -4

f(-4) = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5

Thus, use the (-4, -5) to find the minimum value

Domain and range

f(x) = (x+4)^2 - 5

The domain is the input values shown on the x-axis

The range is the set of possible output values f(x)

Therefore,

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

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2 years ago
How do I solve this question
sammy [17]
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