2W^2 - 11W = -12 add 12 to both sides
2W^2 - 11w + 12 = 0 factor as
(2W - 3) (W - 4) = 0
Set each factor to 0
2W - 3 = 0 W - 4 = 0
add 3 to both sides add 4 to both sides
2W = 3 W = 4
divide both sides by 2
W = 3/2
And those are the two solutions
tokyo we wanted to go to kyoto to shoot some of the video we were making but i hurt my back so we couldnt
sin^5/2(x) → sin^1/2(x)*sin^2(x) so:
sin^1/2(x)(cos(x))-sin^5/2(x)(cos(x)) → sin^1/2(x)*cos(x)*(1 - sin^2(x)) → sin^1/2(x)*cos(x)*cos^2(x) → sin^1/2(x)*cos^3(x)
.
x^2-2y^2-3=0
\(\begin{array}{rcll} x^2-2y^2-3 &=& 0 \qquad | \qquad + 3 \\ x^2-2y^2 &=& 3 \qquad | \qquad : 3 \\ \frac{x^2}{3}-\frac{2y^2}{3} &=& 1 \\ \frac{x^2}{3}-\frac{ y^2}{\frac32} &=& 1 \\ \end{array}\)
parameter notation hyperbola \(\boxed{~ \begin{array}{rcll} \frac{x^2}{3}-\frac{ y^2}{1.5} &=& 1 \\ \end{array} ~}\)
You asked: sin^1/2(x)(cos(x))-sin^5/2(x)(cos(x))=cos^3(x)(sin^1/2(x))
Here's the thing, do you know X?
As follows: \(\frac{1-\cos^2\theta}{\sin\theta} \rightarrow \frac{\sin^2\theta}{\sin\theta} \rightarrow \sin\theta\)
Good Job
No, thats easy...
Here's one for you: How much dirt is in a hole of 3m * 3m *3m?
101
Good job that tricks people did you use a calulater
