We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3)
whole area=A1+A2=25/2+75/2=100/2=
50
The domain of the composite function is given as follows:
[–3, 6) ∪ (6, ∞)
<h3>What is the composite function of f(x) and g(x)?</h3>
The composite function of f(x) and g(x) is given as follows:

In this problem, the functions are:
.
The composite function is of the given functions f(x) and g(x) is:

The square root has to be non-negative, hence the restriction relative to the square root is found as follows:


The denominator cannot be zero, hence the restriction relative to the denominator is found as follows:





Hence, from the restrictions above, of functions f(x), g(x) and the composite function, the domain is:
[–3, 6) ∪ (6, ∞)
More can be learned about composite functions at brainly.com/question/13502804
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Answer:
n+2
Step-by-step explanation:
This is a Typical Question on Arithmetic Progression with initial term as 2n+1
The next number in this particular series will be 2n+1+2 which is 2n+3 and so on. Thus a common difference of 2 exists.
The nth term of an Arithmetic Progression AP, bn is represented below where b1 is the first term
bn=b1 +(n-1)d where d is the common difference
For this particular series, the 4th term b4 is
b4= 2n+1+(4-1)2 0r b4 =2n+1+6=2n+7
sum of series is represented by equation: n(b1 +b4)/2 where b1 and b4 are 1st and 4th term respectively
Thus Sum required =4(2n+1+2n+7)/2=4(4n+8)/2=8n+16 0r n+2