It’s the second one it’s right
R(t) = 4t
A(r) = π(r^2)
a) A(t) = A[r(t)] = π[r(t)]^2 = π[4t]^2 = 16π(t^2)
b) t = 4,
A(4) = 16*3.14*(16)^2 = 12,861.44
Answer:
The sweat chloride reference value is less than 30 mmol/L. A value of more than 60 mmol/L of chloride in the sweat is consistent with a diagnosis of cystic fibrosis. The values of 30-60 mEq/L may represent heterozygous carriers, these carriers cannot be accurately identified with a sweat chloride test.
Answer:
-64
Step-by-step explanation:
5(x – 6) + 3x – 2
Distribute
5x - 30 +3x -2
Combine like terms
8x -32
Let x = -4
8*-4 -32
-32 -32
-64
