49pi is area and 14pi is circumference, look up “area of a circle” in google and a calculator will pop up for you, all you do is type in the info and it’ll use the formula to find the answer
Absolute value cannot be less than 0
Solve Absolute Value.<span><span>|<span>x−5</span>|</span>=<span>−1</span></span>No solutions.
<span>|<span><span>−6</span>−<span>2x</span></span>|</span>=8
<span>x=<span>−<span><span>7<span> or </span></span>x</span></span></span>=<span>1
</span>
<span>|<span><span>5x</span>+10</span>|</span>=10
<span>x=<span><span>0<span> or </span></span>x</span></span>=<span>−4</span>
<span>|<span><span>−<span>6x</span></span>+3</span>|</span>=<span>0
</span>
So your answer is D) |–6x + 3| = 0
Answer: Second option, third option and fifth option.
Step-by-step explanation:
In order to solve this exercise it is important to remember the multiplication of signs:

Knowing that, you can distribute the sign and add the like terms. Repeat this procedure in each expression given in the exercise.
Therefore, you get:

You can identify that the expressions
,
and
are equivalent after simplifying.
Answer:
Step-by-step explanation:
We are given that

Function f decreases from quadrant 2 to quadrant 1 and approaches y=0
It cut the y- axis at (0,6) and passing through the point (1,2).
Function g(x) approaches y=0 in quadrant 2 and increases into quadrant 1.
It passing through the point (-1,2) and cut the y-axis at point (0,6).
Reflection across y- axis:
Rule of transformation is given by

Using the rule then we get

By using

Substitute x=-1

Substitute x=0

Therefore,
is true.
Answer:
Coordinates of Vertices of triangle ABC are A (-4,3) , B(4,4) and C(1,1).
As, DO is Dilation of Δ ABC by Scale factor of
.
Vertices of A' B'C' are

So, Image Δ A'B'C' will be smaller than the Pre image Δ ABC.
The two triangles will be congruent.
AO is Dilated by a factor of half , so A'O' will be half of AO.
So, correct Statements are
1. AB is parallel to A'B'.
2.DO,1/2(x, y) =
The distance from A' to the origin is half the distance from A to the origin.