ma + mb = m(a + b)
this is an algebra formule,
what cyfers, NUMBERS STAND A and M for
I suspect that slashes representing fractions are missing:
m/a + m/b = mb + ma/ab
= m(b + a)/ab
It is mA + mB
Assume a triangle ABC with a line AB (containing the side AB) with external angle D which is formed when line AB and line segment AC intersect. We are asked to prove that the external angle D is equal to the sum of the two interior angles B and C. Angles A and D are supplementary angles (they sum to 180 degrees) because they are linear angles (both together make a straight line, or a 180 degree angle). This means: m<A + m<D = 180 degrees. m<A = 180 deg - m<D Then because A, B, and C are the three angles in a triange: m<A + m<B + m<C = 180 deg m<A = 180 deg - m<B - m<C By substituting 180 deg - m<D in for m<A in the above equation we get: 180 deg - m<D = 180 deg - m<B - m<C Subtract 180 deg from each side: -m<D = -m<B - m<C Multiply both sides by -1 m<D = m<B + m<C Which proves that the measure of the external angle D is equal to the sum of the two opposite interior angles B and C for any given triangle. wow. that's a lot. lol.
To solve the sum and difference of two terms, you can use the identities for the sum and difference of squares. For two terms (a) and (b), the sum is expressed as (a + b) and the difference as (a - b). To find their product, you use the formula: ((a + b)(a - b) = a^2 - b^2). This allows you to calculate the difference of squares directly from the sum and difference of the terms.
(b+5)
Assuming that a and b are two non-negative numbers, then their sum is a + b and the difference is |a - b|.
It is mA + mB
Sum = 38 + M
Assume a triangle ABC with a line AB (containing the side AB) with external angle D which is formed when line AB and line segment AC intersect. We are asked to prove that the external angle D is equal to the sum of the two interior angles B and C. Angles A and D are supplementary angles (they sum to 180 degrees) because they are linear angles (both together make a straight line, or a 180 degree angle). This means: m<A + m<D = 180 degrees. m<A = 180 deg - m<D Then because A, B, and C are the three angles in a triange: m<A + m<B + m<C = 180 deg m<A = 180 deg - m<B - m<C By substituting 180 deg - m<D in for m<A in the above equation we get: 180 deg - m<D = 180 deg - m<B - m<C Subtract 180 deg from each side: -m<D = -m<B - m<C Multiply both sides by -1 m<D = m<B + m<C Which proves that the measure of the external angle D is equal to the sum of the two opposite interior angles B and C for any given triangle. wow. that's a lot. lol.
the sum of 3 times m and n
The sum of "m" and 6 is "m + 6".
Possible. Example: void mat_mul (int m, int n, int l, const int **a, const int **b, int **c) { int i, j, k; double sum; for (i=0; i<m; ++i) { for (j=0; j<l; ++j) { sum= 0; for (k=0; k<n; ++k) { sum += a[i][k] * b[k][j]; } c[i][j]= sum; } }
The sum of m and 6 is written as m + 6. In mathematical terms, this means adding the value of m to 6. For example, if m = 4, then the sum would be 4 + 6, which equals 10. So, the sum of m and 6 is m + 6.
We get a system of equations: a+b=12 a2+b2=90. Replace 12-a for b, and we get: a2+(12-a)2=90 2a2-24a+144=90 2a2-24a+54=0 a2-12a+27=0 a1=9 a2=3 Because of symmetry, we get two equivalent solutions: a=9, b=3 or a=3,b=9
int mul (int a, int b) { int sum= 0; for (; b>0; --b) sum -= -a; for (; b<0; ++b) sum -= a; return sum; }
The sum of is the total of everything being summed; the sum total. Thus the sum of a, b and c is therefore a + b + c.
the sum of m and 9 is 9+m
A number a is even if there exists an integer n such that a = 2n A number b is odd if there exists an integer m such that b = 2m + 1. So: a+b = (2n) + (2m +1) = 2 (n+m) + 1 Since n and m are integers, n+m is also an integer. So a+b satisfies the definition of an odd number.