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Distribute the 49 and 81 first

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Nabeel Gulzar

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Q: How do i factorise 49(m 2n)2 81(2m n)2?
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Related questions

Could you Balance NH4NO3 - N2 O2 H2O?

2NH4NO3->2N2+O2+4H2O


What is the balance to N2 plus O2 equals NO?

The balance is 2N2 +2O2= 4NO


Can you add 2n2-n plus 5 and n2 plus 1?

Yes I can.


How do you balance N2 plus O2 equals N2 O5?

To balance the chemical equation N2 + O2 = N2O5, start by counting the atoms of each element on both sides. You would need to put a coefficient of 2 in front of N2 to balance the nitrogen atoms. Then, add a coefficient of 5 in front of O2 to balance the oxygen atoms. The balanced equation is 2N2 + 5O2 = 2N2O5.


What is the balance equation for N2 and O2?

The balanced equation for the reaction between nitrogen (N2) and oxygen (O2) to form nitrogen dioxide (NO2) is: 2N2 + 4O2 -> 4NO2


What is the factor or 2n2 plus 6n minus108?

2(n2+3n-54) 2(n+9)(n-6)


What is the answer to the sum of the squares of two consecutive positive integers is 3785 find the two integers?

The integers are 43 and 44. The answer can be found by representing the two consecutive numbers as "n" and "n+1". n2 + (n+1)2 = 3785 n2 + (n2 + 2n + 1) = 3785 2n2 + 2n + 1 = 3785 2n2 + 2n - 3784 = 0 n2 + n - 1892 = 0 (n + 44) (n-43) = 0 n = -44, +43 The answer is restricted to positive values, so n = 43 and n+1 = 44


What is the balanced equation if Nitrogen gas and oxygen gas react to form nitric oxide?

The balanced equation for the reaction of nitrogen gas (N2) and oxygen gas (O2) to form nitric oxide (NO) is: 2N2 + O2 → 2NO


How do you work out the nth term of 1 - 5 - 13?

The sequence is too short to be certain but it could be tn = (n-1)2 + n2 = 2n2 -2n + 1


How can you balance NH4NO2 yields N2 H2O?

2nh4no3 ---> 2n2 + o2 + 4h2o


What do you know about the sums and the product of odd and even numbers?

Even numbers : they can be written 2n 2n1+2n2+....+2nm = 2(n1+n2+....+nm) so it's always an even number 2n1x 2n2x....x2nm = 2(n1+n2+....+nm) which is always an even number Odd numbers: they canbe written 2n+1 (2n1+1) + (2n2+1) + ....+ (2nm +1) = 2(n1+n2+....+nm) + m which is even if m is even and odd if m is odd, so it depend of the number of terms (2n1+1) x (2n2+1) x ....x (2nm +1) is always odd cause the last term of the expansion is always +1 and other terms have at least 2 as factor


Can 9 odd numbers make 50?

An odd number is written 2n+1 with n any number (2n1+1)+(2n2+1)+...+(2n9+1) = 2(n1+n2+...+n9)+9 = 50 is not possible as 2(n1+n2+...+n9) = 41 is not possible with integers