The equation of a circle is written as (x-h)^2 + (y-k)^2 = r^r
H and k are the x and y coordinates of the center of the circle and r is the radius.
You are given the diameter coordinates so find the halfway point for the center then calculate the radius
Midpoint = (x1 +x2)/2, (y1 + y2)/2
Midpoint = (7 + -1)/2, (-3 +7)/2
Midpoint = 6/2, 4/2
Midpoint = 3,2
So h = 3 and k = 2
Now find radius by finding the distance between the center point and an end point.
Distance = sqrt(41)
Equation of the circle:
(X-3)^2 + (y-2)^2 = 41
Which one is problem number 4?
Answer:
A n B = {6}
Step-by-step explanation:
Given he sets A = (even numbers less than 10) and B = (multiples of 3 greater than 3)
A = {2, 4, 6, 8}
B = {3, 6, 9, 12, 15...}
A n B includes the numbers are are common to both sets i.e number that we can find in both A and B.
From the set given, we can see that the only number that exists in both set is 6, hence;
A n B = {6}
93% of 149
10% of 149 = 14,9
93% of 149 > 90% of 149
93% of 149 > 149 - 14,9 ≈ 149-15≈134
> 134
⇒ the answer is D
Answer:
If problem is "-|2x+3|+8=12"
then there is (<em>no solution.)</em>
Step-by-step explanation:
Let's solve your equation step-by-step.
−|2x+3|+8=12
Step 1: Add -8 to both sides.
−|2x+3|+8+−8=12+−8
−|2x+3|=4
Step 2: Divide both sides by -1.
−|2x+3|=4
-1 -1
|2x+3|=−4
Step 3: Solve Absolute Value.
|2x+3|=−4
No solutions Because absolute value is less than 0.
