The area is that of two 20 yd squares and one 20 yd circle.
.. A = 2*(20 yd)^2 +(π/4)*(20 yd)^2
.. = (2 +π/4)*(400 yd^2)
.. = (800 +100π) yd^2
.. ≈ 1114.16 yd^2
The perimeter is that of a 20 yd circle and 80 yd more.
.. P = π*20 yd + 80 yd
.. ≈ 142.83 yd
Let's solve your system by substitution.
y
=
−
2
x
+
7
;
y
=
5
x
−
7
Step: Solve
y
=
−
2
x
+
7
for y:
y
=
−
2
x
+
7
Step: Substitute
−
2
x
+
7
for
y
in
y
=
5
x
−
7
:
y
=
5
x
−
7
−
2
x
+
7
=
5
x
−
7
−
2
x
+
7
+
−
5
x
=
5
x
−
7
+
−
5
x
(Add -5x to both sides)
−
7
x
+
7
=
−
7
−
7
x
+
7
+
−
7
=
−
7
+
−
7
(Add -7 to both sides)
−
7
x
=
−
14
−
7
x
−
7
=
−
14
−
7
(Divide both sides by -7)
x
=
2
Step: Substitute
2
for
x
in
y
=
−
2
x
+
7
:
y
=
−
2
x
+
7
y
=
(
−
2
)
(
2
)
+
7
y
=
3
(Simplify both sides of the equation)
Answer: x=2 and y=3
Set up the following equation for this segment:
x is segment AB's length, and 3x is segment BC's length. 20 is segment AC's length.
Combine like terms:
Divide both sides by 4 to get x by itself:
x will equal 5.
Plug this value into the values for both segments:
Segment AB:
Segment AB is 5 inches long.
Segment BC:
Segment BC is 15 inches long.
Answer:
100%
Step-by-step explanation:
