As a product of its prime factors in exponents: 23*32*53 = 9000
3 x 3 x 1000
The answer is 400 because when you take three numbers whose product is 9000, one of the possible sets is 10, 30, and 30. When you multiply these three numbers (10 × 30 × 30), the product equals 9000. If you are looking for a specific scenario where the result is 400, you may be referring to a different interpretation or context of the problem. Could you please clarify the conditions or constraints of the question?
20.800838230519041145300568243579 * 20.800838230519041145300568243579 * 20.800838230519041145300568243579 = 9000
(2) x (3) x (1,500.1) will do it.
9000 = 9000. No other number equals 9000.
300 x 300
9000x1 1x9000 900x1
They are 1,2, and 4,500
3 x 3 x 1000
The answer is 400 because when you take three numbers whose product is 9000, one of the possible sets is 10, 30, and 30. When you multiply these three numbers (10 × 30 × 30), the product equals 9000. If you are looking for a specific scenario where the result is 400, you may be referring to a different interpretation or context of the problem. Could you please clarify the conditions or constraints of the question?
20.800838230519041145300568243579 * 20.800838230519041145300568243579 * 20.800838230519041145300568243579 = 9000
9000 to the nearest thousand, 10000 to the nearest ten thousands and 0 to the nearest million.
Prime factorisation of 9000 = 2*2*2*3*3*5*5*5 Well 9000 can be expressed as the product of (2*2*2)*(3*3)*(5*5*5) or 8*9*125 So, 9000 can be expressed as the product of 8,9 and 125. It can be expressed as 2*(2*2*5*5)*(3*3*5) i.e. 9000 = 2*100*45 By using the method of Prime factorisation, we can find many cases in which product of 3 numbers is 9000. We can also express it as 900*(1/10)*100, which cannot be done using P.F. of 9000. We can find a large number of cases in which product of three numbers is 9000. We can find numbers by this method: Assume any three variables and put their product equal to 9000 i.e. x*y*z = 9000 where x,y and z are real numbers. Put any value of x, say 34 and any value of y, say 23, then on putting the values we can find the value of z i.e. 34*23*z = 9000 782*z = 9000 z = 11.5089 So, this method works better than Prime Factorisation.
(2) x (3) x (1,500.1) will do it.
The product is: 81,000,000
9000 = 9000. No other number equals 9000.
7*9000 = 63000