Answer:
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Kent has two similar cylindrical pipes, pipe a and pipe b. the radius of pipe a is 6 cm, and the radius of pipe b is 2 cm. what
1 answer:
The ratio of the volume of pipe a to the volume of pipe b is 27:1 .
<h3>What is cylinder?</h3>A cylinder is a three-dimensional shape consisting of two parallel circular bases, joined by a curved surface. The center of the circular bases overlaps each other to form a right cylinder.
Given that,
Radius of pipe a is 6 cm, and the radius of pipe b is 2 cm.
We know that the volume of cylinder is :-
where r is radius and h is height of the cylinder.
If two figures are similar then ratio of volume is equal to the cube of any dimension.
The ratio of the volume of Pipe a to the volume of Pipe b :-
=
Hence, The ratio of the volume of pipe a to the volume of pipe b is 27:1 .
To learn more about cylinder from the given link:
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Both the options that Nadia gives are wrong about the length of hypotenuse.
Given length of hypotenuse is 73 and options which are :
1) Because the pythagoras theorem states that squared = 73 squared.
2) 3 squares are positioned to form a right triangle.
We have to choose the correct option which shows the calculation of length of hypotenuse.
Pythagoras theorem says that in a right angled triangle the square of hypotenuse of triangle is equal to sum of the square of base and square of perpendicular.
First option says that
784+2025=5329
2809≠5329
so it is wrong.
Second option says that the length of hypotenuse is the sum of three squares but we have seen that the square of the length of hypotenuse is equal to sum of two square one is the square of base and the square of perpendicular.
Hence both the options are incorrect about the length of hypotenuse.
Learn more about pythagoras theorem at brainly.com/question/343682
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Answer with Step-by-step explanation:
We are given that
for
We have to prove that for all n equal and greater than 1
proof: Let P(n)= for all n equals to and greater than 1
We prove that for all n there exists Z,if n greater and equals to zero
Base case : The base case is n=1
Substitute n=1 then we get
P(1)=-3(1)+5=2
This is true for n=1
Induction step : suppose that P(n) is true for n=k where k greater than 1
Then P(k)=-3k+5
Now, we shall prove that for n=k+1, P(n) is true for n=k+1
P(k+1)=-3(k+1)+5
P(k+1)=-3k-3+5=-3k+5-3
We are given that for n equal to and greater than 2
Therefore, P(n) is true for n=k+1
Hence, P(n)=-3n+5 for all n equal and greater than 1