Factor to this equation out, the easiest way is to solve for the two x values that add up to -1 but when multiplied is equal to -6. It should look like this: (x +- ?)(x +- ?).
The factored out form is then (x - 3)(x +2), because -3 + 2 = -1 and when multiplied together gives -6.
x - 3 = 0
x = 3
x + 2 = 0
x = -2
The two solutions are therefore, x = 3 and x = -2.
Hope it helps
Answer:

Step-by-step explanation:
Let 
![m=(y^3)^{\frac{1}{2}}\\\\m=y^{3\times \frac{1}{2}}\ \ \ \ \ \ \ \ \ [as\ (x^a)^b=x^{ab}]\\\\m=y^{\frac{3}{2}](https://tex.z-dn.net/?f=m%3D%28y%5E3%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cm%3Dy%5E%7B3%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5Bas%5C%20%28x%5Ea%29%5Eb%3Dx%5E%7Bab%7D%5D%5C%5C%5C%5Cm%3Dy%5E%7B%5Cfrac%7B3%7D%7B2%7D)
No because then they would be different triangles. Congruent triangles are triangles that are the same
I hope I've helped!
The answer to this equation would be 3.
Answer:

Step-by-step explanation:
Given:
The above triangle
Required
Solve for AB in terms of a, b and angle C
Considering right angled triangle BOC where O is the point between b-x and x
From BOC, we have that:

Make h the subject:

Also, in BOC (Using Pythagoras)

Make
the subject

Substitute
for h
becomes


Factorize

In trigonometry:

So, we have that:

Take square roots of both sides

In triangle BOA, applying Pythagoras theorem, we have that:

Open bracket

Substitute
and
in 


Open Bracket

Reorder

Factorize:

In trigonometry:

So, we have that:


Take square roots of both sides
