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.
X to the 7th power. X^m*X^n=X^m+n That means when you multiply variables with the same base, you add the exponents.
am * an = am+n
m^4 n^5 - m^20 n^21
the sum of 3 times m and n
ab*ac=ab+c consider the powers of 2. 22=4, 23=8, 22*23=32=23+2=25 when multiplying a number by itself, you raise its power by one. when multiplying a number by itself n times, you raise it to the power of n, so if you raise a number to the power n, then the seame number to the power m, then multiply these together you are multiplying n+m times
[(4/n)(9)(2/9)]^n -2x^6 - 2n=m/x^2 (8/n)^2 - 2x^6 -2n=m/x^2 (64x^2)/n^2 -2x^8 -2nx^2=m Now we know what m equals. I've got to go now. Sorry!
If n is a natural number and M is a matrix, then Mn denotes the matrix M multiplied by itself n times. We can include n=0, but that is just the identity matrix. So the power of a matrix is very similar to the exponents that are used for numbers.
for any non zero no. x, x^0=1 the proof is as follows, consider the two no.s x^m and x^n,where m and n are two non zero no.s. now let us assume without any oss of generality,that m>n,hence (x^m)/x^n=(x*x*x....m times)/(x*x*x...n times) now on the r.h.s, n no. of x in the denominator will cancel out n no. of x in the numerator(as x is non zero);leaving (m-n) no. of x in the numerator, i.e. (x^m)/(x^n)=x^(m-n) now letting m=n,we have x^m/x^m=x^(m-m) or, 1=x^0 hence the proof if x is also 0,i.e. 0 to the power 0 is undefined!
m and n are 70 and 90
If we are to find the product of 5 and m and n/2 (which is half of n), we have: 5 times m times n/2 = 5 x m x n/2 = 5mn/2
P = 1 For K = 1 to M . P = P * N Next K PRINT "N raised to the power of M is "; P
m -5 n-3
x^a is x time itself a number of timesx^0=1 any number raised to the zero power equals 1x^n times x^m = x^(n+m)x^n divided by x^m = x^(n-m)x^(1/n) is the n-th root of xx^(-1) is 1/x(mathematics) Any of the laws aman am+n, am/an am-n, (am)n amn, (ab)n anbn, (a/b)n = an/bn; these laws are valid when m and n are any integers, or when a and b are positive and m and n are any real numbers. Also known as exponential law.
1 Watt = 1 N-m/sec 1800 N-m/sec = 1800 Watts
For a positive number, n, x raised to the power n is 1 multiplied by x n times. It is often wrongly described as x multiplied by itself n times.Thus, x^3 = x*x*x [ or 1*x*x*x] but notx multiplied by x multiplied by x multiplied by x which would be x^4.For negative integers, m, x^m = (1/x)^m = 1/(x^m)x^0 = 1 for all non-zero x.For fractional powers,x^(m/n) = the nth root of x^m or, equivalently, (nth root of x)^m.
They are named as K , L , M , N ... . Where in K is the first shell , L is the second shell , M is the third shell , N is the fourth shell and so on.
3n^4 divided by 3n^3 = n
For the purposes of this answer we will use the convention that x^n means xⁿ Any non-zero number to the zero power is 1 we get this from the idea that (x^n)(x^m) is x^(n+m) if n=-m, then n+m = 0 and x^ m is 1/(x^n) and (x^n)(x^m) (x^n)/(x^n)=1 so x^0 = 1 1 raised to any positive power is 1 1 raised to any negative power is 1 We can thus represent 1 as x^0 = 1 1^n = 1 1 ^(-n) = 1 (where n>0 Note that n need not be an integer since all roots of 1 are also 1)
It is m7n-7.
Either it is a prime or put m = n! Is that it?
Where m and n are statements m n is called the _____ of m and n.
There's no such thing as "the fifth square" of a number.To calculate any whole-number power of a number, write the number that many timesall on one line, then write the word "times" between them, get your pencil out, and getto work.The 5th power of a number 'N' is'N' times 'N' times 'N' times 'N' times 'N' .