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erastovalidia [21]
3 years ago
9

If the radius of the base of the cone is 8/3 units and the height is 12 units, what is the volume?

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

89.32

Step-by-step explanation:

The volume of a cone with radius 8/3 and height 12 is

V= \frac{1}{3}\pi r^2 h=  \frac{1}{3}\pi (\frac{8}{3}^2)(12) =  \frac{1}{3}\pi (\frac{64}{9})(12)= \frac{1*64*12}{3*9}\pi = \frac{256}{9} \pi = 89.32.

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For f(x) = 3x - 1 and g(x) = x+2, find (f – g)(x).
kobusy [5.1K]
Well, I bet you want your answer right away! So here it is.


<span>Given <span>f (x) = 3x + 2</span> and <span>g(x) = 4 – 5x</span>, find <span>(f + g)(x), (f – g)(x), (f × g)(x)</span>, and <span>(f / g)(x)</span>.</span>

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f  × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

<span>\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}<span><span>(<span><span>​g</span>​<span>​f</span><span>​​</span></span>)</span>(x)=<span><span>​<span>g(x)</span></span>​<span>​<span>f(x)</span></span><span>​​</span></span></span></span><span>= \small{\dfrac{3x+2}{4-5x}}<span>=<span><span>​<span>4−5x</span></span>​<span>​<span>3x+2</span></span><span>​​</span></span></span></span>

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f  × g ) (x) = –15x2 + 2x + 8

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Hope I helped! :) If I did not help that's okay.


-Duolingo
</span></span></span></span>

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