It is: 5*7 = 35 square m
If: 9m = 4m-35 Then: m = -7
Centi is a factor of one hundredth; therefore 35 cm is .35m. Add 5m and .35 m is 5.35m.
Let the numbers by 'm' & 'n' Hence m + n = 12 mn = 35 Hence m = 35/n Substitute 35/n + n = 12 Multiply through by 'n' 35 + n^2 = 12n n^2 - 12n + 35 = 0 It is now in Quadratic Form ; Factor ( n - 7)(n - 5) = 0 Hence n= 7 or n= 5 It follows that m = 5 , or m = 7
Therefore 5x=35, x=35/5, hence x=7
It is: 5*7 = 35 square m
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If: 9m = 4m-35 Then: m = -7
It is: 35 square m
Centi is a factor of one hundredth; therefore 35 cm is .35m. Add 5m and .35 m is 5.35m.
Let the numbers by 'm' & 'n' Hence m + n = 12 mn = 35 Hence m = 35/n Substitute 35/n + n = 12 Multiply through by 'n' 35 + n^2 = 12n n^2 - 12n + 35 = 0 It is now in Quadratic Form ; Factor ( n - 7)(n - 5) = 0 Hence n= 7 or n= 5 It follows that m = 5 , or m = 7
What can 5 and 7 both go into
Well, let's think of this like a happy little math problem. When we divide 35 by 6, we get 5 with a remainder of 5. So, we can write this as the mixed number 5 5/6 or as the improper fraction 35/6. Either way, it's just a different way of looking at the same beautiful math problem.
1 m = 100 cm ⇒ 35 cm = 35/100 m = 7/20 m (divide top and bottom by 5)
Father Knows Best - 1954 Bud Has a Problem 5-35 was released on: USA: 11 May 1959
.35 m = 35 cm, and 35 cm > 31 cm. 31 cm = .31 m, and .31 m < .35 m Any way you do the math, .35 m is a greater diameter than 31 cm.
initial problem: 5*m=15 divide both sides by 5: m=(15)/(5) solution: m=3