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3 numbers whose product is about 9000? - Answers
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.
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3 numbers whose product is about 9000? - Answers
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.
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3 numbers whose product is about 9000? - Answers
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.
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- og:descriptionPrime 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.
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