Perimeter of a square = 4s where is the length of one side.
Equation:
Area = s^2
169 = s^2
s = 13 inches
B. 13 INCHES
They did not include the constraint for y ≤x+3 on the graph.
See attached picture with added constraint.
Using the 4 points that are given as the solution on the graph, replace t he x and Y in the original equation to solve and see which is the greater value.
Point (0,3) P = -0 +3(3) = 0+9 = 9
Point (1,4) P = -1 + 3(4) = -1 +12 = 11
Point (0,0) P = -0 + 3(0) = 0 + 0 = 0
Point (3,0) P = -3 + 3(0) = -3 + 0 = -3
The correct solution to maximize P is (1,4)
14. Trick question, ignore the side.
Call the angle b.
34 + b + b = 180
2b = 146
b = 73
choice b
15. That's intersecting chords so the product of the segment lengths are equal:
5(22)=10 x
x = 5(22)/10 = 11
choice a
16. Intersecting tangent and secant;


Choice a
Answer:
P = 4(2x + 5)
P = (2x + 5)+(2x + 5)+(2x + 5)+(2x + 5)
Step-by-step explanation:
Given that:
Perimeter of square = 4 * s
Where s = side length
If perimeter, P = 8x + 20
Then,
4s = 8x + 20
Divide both sides by 4
4s / 4 = (8x + 20) / 4
s = 2x + 5
P = 4s
P = 4(2x + 5)
Or
P = 2x + 5 in 4 places
P = (2x + 5)+(2x + 5)+(2x + 5)+(2x + 5)
The answer is 9/5 as 3×3 is 9 and you are dividing it by 5.