Simplify expressions 2y(y+3)+4y(y-1)
2 answers:
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Answer:
Option A) Inside the circle
Step-by-step explanation:
step 1
Find the radius of the circle
we know that
The radius is equal to the distance from the center to any point on the circle
the formula to calculate the distance between two points is equal to
we have
A(-5,-8),M(-1,-3)
substitute the values
step 2
Find the distance from the center to point V
we know that
If the distance from the center to point V is equal to the radius, then the point V lie on the circle
If the distance from the center to point V is less than the radius, then the point V lie inside the circle
If the distance from the center to point V is greater than the radius, then the point V lie outside the circle
we have
A(-5,-8),V(-11,-6)
substitute in the formula
so
The distance from the center to point V is less than the radius
therefore
The point V lie inside the circle
Hope that helped :)
Answer:
CD ≠ EF
Step-by-step explanation:
Using the distance formula
d =
with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)
CD =
=
= =
Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)
EF =
=
= =
Since ≈
, then CD and EF are not congruent
Here is my process for solving this.
First I drew arrows that indicated I was moving the whole triangle 5 units to the left.
*Look at first attachment*
Then I drew another triangle using those new points. (The new triangle is in pink)
*Look at second attachment*
Then I drew arrows that moved this new triangle 4 units up. (The new arrows are in pink)
*Look at third attachment*
Then I drew the new triangle in blue using the new points.
*Look at fourth attachment*
Then I mirrored / reflected the triangle over the x axis (the horizontal line) In green.
*Look at fifth attachment*
The fifth attachment in green is the final product! Hope that helps.
