Step-by-step explanation:
that r = P/(2π) so A = π(P/(2π))2 = P2/(4π). Any positive area less than this is also possible. So in this problem the largest area possible is (16 in.)2/(4π) = 64/π sq.
Answer:
4 units
according to question:
perimeter of square = 16 units
4a = 16
where a is the side length
a = 16/4
= 4 units
Answer: B
Grade must be greater than or equal to 88
78.5÷3.14=25
56.272×41.2=2318.4064
429×338×712=103241424
447÷513=0.8713450292
hope it will help!
1 2 3 6
1x6=6
2x3=6
3x2=6
6x1=6
