Parallel lines have the same slope
Find the slope of points A and B
(y2 - y1)/ (x2-x1)
(5-(-5))/(-1-(-2)) = (5+5)/(-1+2) = 10/1
The slope of the points is 10
In the equation
y = 10x + 12
Y = mx + b, m indicating slope
The slope of the equation is also 10
Because both slope is 10, the line through the points and the equation is parallel
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Answer:
11.25
Step-by-step explanation:
First, add the values up.
15+9+11+7+5+11+26+6=90
Then divide the sum by the number of numbers.
90/8=11.25
It's either 15, 17 or 18. Anything higher would make angle AC much longer than seen in the image.
ANSWER = c. 63°
EXP:
>ABC = 1/2(x + y)
