That depends on the numerical value of W.
Their product.
That depends on the numerical value of w.
Any multiple of 9w
all the multiples of 36 (because 6 goes into 36 w/o remainder) 36, 72, 108, 144, etc.
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). Voilà!
P of rectangle = 2(L+W) where L=W+6 24 = 2(W+(W+6)) 12 = W+ W+ 6 6 = 2 W 3 = W The length (W+6) is 9, width (W) is 3.
As w and w are the same number, there are no numbers between w and w. However, if it is inclusive of the limits, (ie "between w and w inclusive"), then there is only 1 number w, which to be a multiple of 4 6 and 8 must be a multiple of their lowest common multiple (lcm) which is 24; ie all multiples of 24, namely w is one of: 24, 48, 72, 96, 120, 144, 168, ...
That depends on the numerical value of w.
6
6 -w over 8 = -2
w + 3/2 > 6 Subtract 3/2 from both sides: w > 6- 3/2 = 41/2
2 64 is also 26, so the sixth root of 26 is 2. In general, the n-th root of Xn is X. ALTERNATIVE ANSWER: The Fundamental Theorem of Algebra states that every polynomial of degree n has n (possibly complex) roots. This means that there are 6 sixth roots of 64. These roots are: 2*w, 2*w^2, 2*w^3, 2*w^4, 2 * w^5 and 2 * w^6 where w is the 6th root of unity, or w = e^(2πi/6) = cos(2π / 6) + i * sin (2π / 6) Note that w^3 = -1 and w^6 = 1, so there are actually two real sixth roots: 2 and -2. The other four are imaginary/complex.