A+30 = 60
a = 30
a + 2b = 60
30+2b = 60
2b = 30
b = 15
5b - 5c = 60
5(15) - 5c = 60
5c = 15
c = 3
10c + d = 60
10(3) + d = 60
30 + d = 60
d = 30
2d + 6e = 180 - 60
2(30) + 6e = 120
6e = 60
e = 10
4f + 4e = 120
4f + 4(10) = 120
4f = 80
f = 20
<u>Answer</u>
The first student was right.
The length of the long side is at least 13 inches.
<u>Explanation</u>
The perimeter of any figure is the distance all round.
Perimeter of a rectangle = 2(l+w). Where l is length and w is the width.
2(l + w) ≥ 30
2{(x-3) + 2} ≥ 30
2(x-3+2) = 30
2(x - 1) ≥ 30
x - 1 ≥ 15
x ≥ 16
When x = 16,
l = 16-3
= 13 inches
The length of the long side is 13 inches. The first student was right.
The answer is C 16:20 because both numbers 4&5 are multiplied by four
52m^2= 2[(x-5)+(x)]. 52m^2= 2(2x-5). 52m^2=4x-10
