As a quadratic expression it is: w^2 +10w +21
11 + 5w
Ah, math time, my favorite! To reduce the fraction w^2 + 5w + 6 over w^2 - w - 12, first factor both the numerator and the denominator. The numerator factors into (w + 2)(w + 3), and the denominator factors into (w + 3)(w - 4). Cancel out the common factor of (w + 3) in both the numerator and the denominator, leaving you with (w + 2) over (w - 4). VoilΓ !
36ong +36ong 612ght w=3 r=6 i=12
Written out f(w) = w^2 + 7w + 12 there is no particular answer since it depends on what w is equal to. normally this type of question is posited as "Solve for w with f(w) = 0?" or to change its form to (ax + b)(cx + d) if f(w) = w^2 + 7w + 12 and f(w) = 0 then 0 = w^2 + 7w + 12 simply look at the factors of 12 1 x 12 2 x 6 3 x 4 now consider (w + a)(w + b) = w^2 + aw + bw + ab = w^2 + (a+ b)w + ab since you know 3 x 4 = 12 , and 3 + 4 = 7 this give you 0 = (w + 3) (w + 4) or f(w) = (w + 3)(w + 4) solving for zero w = either -3 or -4 as (-3 + 3)(-3 + 4) = (0)(1) = 0 or (-4 + 3)(-4 + 4) = (-1)(0) = 0 so depending on what you are actually looking for the answer is w^2 + 7w +12 = (w + 3)(w + 4) or -3 and -4
Three plus w.
6 plus w
whats w stand for? -7w + 3(w+2)= -14
As a quadratic expression it is: w^2 +10w +21
W/3 + 2W/3 is one possible answer.
w + 3 = 4w - 6 9 = 3w w = 3
-3
w+3=w+6 obviously doesn't equal itself... subtract w from both sides and you are left with 3=6 So the answer is NO SOLUTION, or as a set, it is the empty or null set.
-4
11 + 5w
6w
(2w + 3)(w + 5)