You would take out a 4x out of both terms because this is a common factor
4x^3-36x
4x(x^2-9)
and then you would notice that (x^2-9) is a difference of squares, so that factors out too
4x(x-3)(x+3)
that is all you can factor! hope this helps
OK so where is the weekly math homework!?
Answer:
Step-by-step explanation:
Problem One
All quadrilaterals have angles that add up to 360 degrees.
Tangents touch the circle in such a way that the radius and the tangent form a right angle at the point of contact.
Solution
x + 115 + 90 + 90 = 360
x + 295 = 360
x + 295 - 295 = 360 - 295
x = 65
Problem Two
From the previous problem, you know that where the 6 and 8 meet is a right angle.
Therefore you can use a^2 + b^2 = c^2
a = 6
b =8
c = ?
6^2 + 8^2 = c^2
c^2 = 36 + 64
c^2 = 100
sqrt(c^2) = sqrt(100)
c = 10
x = 10
Problem 3
No guarantees on this one. I'm not sure how the diagram is set up. I take the 4 to be the length from the bottom of the line marked 10 to the intersect point of the tangent with the circle.
That means that the measurement left is 10 - 4 = 6
x and 6 are both tangents from the upper point of the line marked 10.
Therefore x = 6
If I am correct, it would be dropped at -3, side adding to negative numbers make them increase to positive numbers.
Answer:
B.) uv = 2.24 and u · v = 5.00
Step-by-step explanation:
Right on Edge. :)
