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Paul [167]
3 years ago
10

The formula for finding the volume of a cone is V = 1/3πr2h. The volume of a cone is 300 cm3 and the height of the cone is 10 cm

. What is the approximate radius of the cone?
A) 3 cm
B) 5 cm
C) 9 cm
D) 28 cm
Mathematics
2 answers:
telo118 [61]3 years ago
8 0
If we solve for radius
V=1/3pir^2h
divide both sides by pih
V/(pih)=1/3r^2
times 3 both sides
3V/(pih)=r^2
sqrt both sides
sqrt(3v/(pih))=r

so if h=10
v=300
then
sqrt(3*300/(pi*10))=r
sqrt(900/10pi)=r
sqrt(90/pi)=r
use calculator
5.35=r
about 5
B is answer
hichkok12 [17]3 years ago
4 0
The answer is b) 5



hope I helped!
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Answer:

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Step-by-step explanation:

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If(x) = x + 2 and h(x) = x-1, what is f • h](-3)?
deff fn [24]

Answer/Step-by-step explanation:

Composition functions are functions that combine to make a new function. We use the notation ◦ to denote a composition.

f ◦ g is the composition function that has f composed with g. Be aware though, f ◦ g is not

the same as g ◦ f. (This means that composition is not commutative).

f ◦ g ◦ h is the composition that composes f with g with h.

Since when we combine functions in composition to make a new function, sometimes we

define a function to be the composition of two smaller function. For instance,

h = f ◦ g (1)

h is the function that is made from f composed with g.

For regular functions such as, say:

f(x) = 3x

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What do we end up doing with this function? All we do is plug in various values of x into

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outputs for each input:

f(−2) = 3(−2)2 + 2(−2) + 1 = 12 − 4 + 1 = 9 (3)

f(0) = 3(0)2 + 2(0) + 1 = 1 (4)

f(2) = 3(2)2 + 2(2) + 1 = 12 + 4 + 1 = 17 (5)

When composing functions we do the same thing but instead of plugging in numbers we are

plugging in whole functions. For example let’s look at the following problems below:

Examples

• Find (f ◦ g)(x) for f and g below.

f(x) = 3x + 4 (6)

g(x) = x

2 +

1

x

(7)

When composing functions we always read from right to left. So, first, we will plug x

into g (which is already done) and then g into f. What this means, is that wherever we

see an x in f we will plug in g. That is, g acts as our new variable and we have f(g(x)).

g(x) = x

2 +

1

x

(8)

f(x) = 3x + 4 (9)

f( ) = 3( ) + 4 (10)

f(g(x)) = 3(g(x)) + 4 (11)

f(x

2 +

1

x

) = 3(x

2 +

1

x

) + 4 (12)

f(x

2 +

1

x

) = 3x

2 +

3

x

+ 4 (13)

Thus, (f ◦ g)(x) = f(g(x)) = 3x

2 +

3

x + 4.

Let’s try one more composition but this time with 3 functions. It’ll be exactly the same but

with one extra step.

• Find (f ◦ g ◦ h)(x) given f, g, and h below.

f(x) = 2x (14)

g(x) = x

2 + 2x (15)

h(x) = 2x (16)

(17)

We wish to find f(g(h(x))). We will first find g(h(x)).

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g( ) = ( )2 + 2( ) (19)

g(h(x)) = (h(x))2 + 2(h(x)) (20)

g(2x) = (2x)

2 + 2(2x) (21)

g(2x) = 4x

2 + 4x (22)

Thus g(h(x)) = 4x

2 + 4x. We now wish to find f(g(h(x))).

g(h(x)) = 4x

2 + 4x (23)

f( ) = 2( ) (24)

f(g(h(x))) = 2(g(h(x))) (25)

f(4x

2 + 4x) = 2(4x

2 + 4x) (26)

f(4x

2 + 4x) = 8x

2 + 8x (27)

(28)

Thus (f ◦ g ◦ h)(x) = f(g(h(x))) = 8x

2 + 8x.

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Step-by-step explanation:

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