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What is 1.0n plus 2m?
It is simply n + 2m
What is 2m plus 3n-m plus 4n-3m?
n-2m
What is the answer to -2m to the fourth power times n to the 6thpower to the 2nd power?
The question is open to multiple interpretations but I think you mean [(-2m)^4] x (n^6)^2 = [(-2)^4](m^4)(n^12) = 16(m^4)(n^12) or 16 times m to the 4th power times n to the 12th power.
If two odd numbers are added together what is the result?
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
What are five numbers with a range of 8 a median of 6 and a mode of 4?
The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.
What is the expanded form of 2m -n?
2m-n cannot be expanded.
What is Factor 0.04m2 - n2?
.04m2-n2=(.2m+n)(.2m-n)
2m - n - m - 2n?
0
What is 1.0n plus 2m?
It is simply n + 2m
The sum of 25 consecutive integers is divisible by?
The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!
How do you find equation for a pyramid of oranges stacked 1 2x2 3x3 4x4?
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
What is 2m divided by 5?
0.4
What is 2m plus 3n-m plus 4n-3m?
n-2m
What is an algebraic expression for the verbal phrase the quotient of a number plus 6 and 3?
6+3*n
What is the answer to -2m to the fourth power times n to the 6thpower to the 2nd power?
The question is open to multiple interpretations but I think you mean [(-2m)^4] x (n^6)^2 = [(-2)^4](m^4)(n^12) = 16(m^4)(n^12) or 16 times m to the 4th power times n to the 12th power.
How do you solve L equals 2M plus 2N for N?
N=l-m
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
