How do I identify two whole numbers
1 answer:
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Answer: A
Step-by-step explanation:
The line 'd' separates the square into two triangles. Since the side lengths are equal, the angle is 45°. We need the line separating the two 90° angles, that I am going to define as 'z'. We can use the sine function.
z = 5
To solve for x, we can use sine again but for the 30 degree angle.
Answer:
(x-12)(x+12) and (2a+b)(x-3y)
Step-by-step explanation:
So first we have:
x^2-144
The key here is that there is no bx in the ax^2+bx+c equation here.
This means that b must equal 0, or in other words, cancel itself out.
To do this, first, the two factors must have opposite signs, and secondly, the blank values must be equal (x+_) * (x-_)
What could they be? The c value, 144, equals the two blank values multiplied together. Since they are in the same in this case, blank value^2 = 144, or
So the answer is:
(x+12)(x-12)
Next we have:
2ax - 6ay + bx - 3by
This is a bit more tricky, but first we must facor out some numbers/variables to simplify.
Lets seperate the two parts of this equation:
2ax - 6ay and bx - 3by
Now lets look at:
2ax - 6ay
How can we simplify this? Factor the 2a, which both have in common:
2a(x-3y)
Now lets look at and factor it too:
bx - 3by
Only b can be factored out, which gives us:
b(x-3y)
Now lets put this back into our orginal equation:
2a(x-3y) + b(x-3y)
Both of these have x-3y. Concidence? Nope!
From the way we split the equation into two pieces and factored, the two x-3ys are actually the same, and we can combine them!
Then, the 2a and the +b can be combined to give us:
(2a+b)
Puttting these together gives us:
(2a+b)(x-3y)
Hope this helps!