m(m - 6) = 91 m(m) + m(-6) = 91 m2 - 6m = 91 subtract 91 from both sides m2 - 6m - 91 = 0 since -91 = (-13)(7) and -13 + 7 = -6, then (m + 7)(m - 13) = 0 m + 7 = 0 or m - 13 = 0 m = -7 or m = 13
The set of integers is closed under addition.Addition is commutative and associative. There exists a unique number, 0, such that n + 0 = 0 + n for any integer n. For every integer m, there exists an integer m' such that m + m' = m' + m = 0. m' is denoted by "-m".
It is: 0 times 10 = 0
10293847560192837465 times 0= 0
m6 - 16m3 + 64 = 0 m6 - 8m3 - 8m3 + 64 = 0 m3(m3 - 8) - 8(m3 - 8) = 0 (m3 - 8)2 = 0 (m3 - 2m2 + 2m2 - 4m + 4m - 8)2 = 0 [m2(m-2) + 2m(m-2) + 4(m-2)]2 = 0 [(m2 +2m + 4)(m-2)]2 = 0 So m -2 = 0 or m2 + 2m + 4 = 0 ie m= 2 or m= -1 +/- i*sqrt(3) with each of the three roots being double roots.
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It's gonna be -1. (2-m)*0 + (-1) = 0-1 = -1. Anything multiplied by 0 is 0 so that takes care of the (2-m)*0 and -1 is just -1. So -1 is the answer.
m(m - 6) = 91 m(m) + m(-6) = 91 m2 - 6m = 91 subtract 91 from both sides m2 - 6m - 91 = 0 since -91 = (-13)(7) and -13 + 7 = -6, then (m + 7)(m - 13) = 0 m + 7 = 0 or m - 13 = 0 m = -7 or m = 13
M=0 n=0 m*n=0
Answer: 'm' times. Examples: If m=2, Then m2 = (2x2) = 4 -> 4-2-2 = 0 (So subtracted TWICE) If m=4 Then m2 = (4x4) = 16 -> 16-4-4-4-4 = 0 (So subtracted FOUR TIMES) This works no matter how large the number is!
Well, hello there! When we see "m to the third power," it means we are multiplying 'm' by itself three times. So, if we have m x m x m, that's what m to the third power represents. It's like painting a lovely little trio of 'm's dancing together on our canvas of mathematics.
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!
If: -6+m = 0 Then: m = 6
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The set of integers is closed under addition.Addition is commutative and associative. There exists a unique number, 0, such that n + 0 = 0 + n for any integer n. For every integer m, there exists an integer m' such that m + m' = m' + m = 0. m' is denoted by "-m".
3m^2 - 9m = 0 factor out 3m 3m(m - 3) = 0 m = 0 -------- or m = 3 --------
0 = 2m^2 - 2m + 36 0 = 2(m^2 - m + 18) 0 = m^2 - m + 18