MATHEMATICS HIGH SCHOOL

A junior basketball has a diameter of approximately 7 in., and a regulation basketball has a diameter of approximately 9.5 in. about how many times as great is the volume of the regulation basketball as the volume of the junior basketball?

Answers

Answer 1
Answer: Scale factor  = 9.5 / 7
the ratio of the volumes will be 9.5^3 / 7^3

=  2.5  to nearest thousandth

So the volume the regulation ball is  2.5 times the volume of the junior one.

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MIDDLE SCHOOL

what must be subtracted from 4x4 - 2x3 - 6x2 + x - 5 so that the answer is exactly divisible by x + x - 2

Answers

To solve this problem, what we have to do is to divide the whole equation 4 x^4 – 2 x^3 – 6 x^2 + x – 5 with the equation 2 x^2 + x – 1. Whatever remainder we get must be the value that we have to subtract from the main equation 4 x^4 – 2 x^3 – 6 x^2 + x – 5 for it to be exactly divisible by 2 x^2 + x – 1.

By using any method, I used long division we get a remainder of -6.

Therefore we have to subtract -6 from the main equation which results in:

4 x^4 – 2 x^3 – 6 x^2 + x – 5 – (-6) = 4 x^4 – 2 x^3 – 6 x^2 + x + 1

MIDDLE SCHOOL

What is y-2=-2.5(x-1) in STANDARD FORM? (It has to be in Ax+By=C form)

Answers

Y-2=-2.5(x-1)
Y-2=-2.5x+2.5
2.5x+y=4.5
COLLEGE

Make a conjecture about the sum of the first 15 positive even number

Answers

I conclude that the sum will be even because any even number can be represented by 2n where n is a whole number
and even numbers are 2 apart, so
the sum of the first 15 are
2n+2(n+1)+2(n+2) etc until we get to 2(n+14)
we can undistribute the 2 from all of them and get
2(n+n+1+n+2...n+14)
and we are sure that whatever is in the parenthasees is a whole number because whole+whole=whole
therefor, the sum is even

if you did want to find the sum then
an=2n
the 15th even number is 30
the first is 2
S15=15(2+30)/2=15(32)/2=15(16)=240
which is even
HIGH SCHOOL

Two arcades have different pricing structures. Arcade A costs $5 to enter and $0.60 per game. Arcade B costs $7 to enter and $0.40 per game. How many games do you have to play in order for the two arcades to be the same price?

Answers

Answer:

you would have to play 10 games

Answer:

10 games

Step-by-step explanation:

Let C be cost and x be number of games

Arcade A: C = 5 + 0.6x

Arcade B: C = 7 + 0.4x

for both arcades to be the same price, Both C's must be equal, hence we equate the right side of both equations:

5 + 0.6x = 7 + 0.4x   (subtract 5 fromboth sides)

0.6x = 7 + 0.4x - 5

0.6x = 2 + 0.4x  (subtract 0.4x from both sides)

0.6x - 0.4x = 2

0.2x = 2 (divide both sides by 0.2)

x = 2 / (0.2)

x = 10

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