30
-5
seventeen over fifteen because the common denominator of 5 and 3 is 15 think about it like this: 4/5x3, 4x3=12, 5x3=15 1/3x5, 1x5=5, 3x5=15 so you get 12/15+5/15 see how they simplify back? then add the numerators and keep the denominator and you get 17/15 or 1 and 2/15
No. You can only add matrices of the same size.
Its called the product. For example, in the problem 5x3=15, 15 is the product.
The GCF is 5x
Since 5x is a factor of both terms, divide it. 5x3 + 5x = 5x(x2 + 1)
The answer is 21
3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
5(x3 - 3x2 - 238)
It is a quadratic expression
If the original is 2x + 3x2 + 6 + 5x3, you should rearrange the terms in order of exponent:5x3 + 3x2 + 2x + 6
If that's 9a2, the answer is a(a2 - 9a + 3) x(5x2 + 3x - 1) -8d(d2 + 3d + 1)
The GCF is 5x3
Yes, 5x3 is the same as 5+5+5
Commutivity of multiplication or the commutative property of multiplication.
5x3(x + 3)(x - 3)(x2 + 9)