Answer:
Step-by-step explanation:
The point of this question is to find out the point where two lines intersect. First we need to get the equation of those lines
Slope of line 1:
(Yb -Ya)/(Xb - Xa) =
(-10 - (-14))/(-1 - (-3)) =
4/2 =
2
Use that slope to find the Y-intercept of line 1
y = 2x + b
-14 = 2(-3) +b
-14 = -6 + b
-8 = b
Therefore Line 1 is:
y = 2x - 8
Slope of line 2
(11 - 13)/(-1 - (-3)) =
-2/2 =
-1
Y-intercept of line 2
y = -x + b
13 = -(-3) +b
13 = 3 + b
10 = b
Therefore line 2 is
y = -x + 10
Now we have 2 equations to solve for the coordinates x and y
y = 2x - 8
y = -x + 10
Substitute y out in one of the equations
2x - 8 = -x + 10
3x = 18
x = 6
Plug x into one of the equations
y = 2(6) - 8
y = 12 - 8
y = 4
Therefore the solution is:
x=6, y=4
The area of the square is 81 sq in, and the area of the circule is (3.14)(3 in)^2, or approx 3.14(9) sq in, or approx 28.27 sq in.
Subtract: 81 sq in - 28.27 sq in = approx. 52.73 sq in.
The difference is 52.73 sq in; the square is larger, the circle smaller.
B is correct.
A. 5(x-2)= 5x-10
B. 4(x+3)= 4x + 12
C. 3(x+1)= 3x+3
D. 2(x+7) = 2x + 14
one hundred and forty four point one four four.
E+r=
(e=9)
r=1.50)
or 9+1.50=
I think
