<h3>Given</h3>
S = πr√(r^2+h^2)
h = 8 m (constant)
<h3>Find</h3>
An approximation of S when r changes from 9 to 8.9
<h3>Solution</h3>
Such an approximation is usually made by estimating the change using the first derivative. That derivative with respect to r is
... S' = π√(r^2+h^2) + πr(1/2·r)/√(r^2+h^2)
... S' = π(2r^2 +h^2)/√(r^2 +h^2) . . . . . use a common denominator
For r=9, h=8, this is
... S' = π(2·81 +64)/√(81+64) = 226π/√145 ≈ 58.96
Then the change in lateral surface area will be approximately
... ∆S ≈ (∆r)·S' ≈ (-0.1)·(58.96) ≈ -5.90 . . . m²
Answer:
I'm not smart but I think there's no solution
Answer:
$44
Step-by-step explanation:
Hello from MrBillDoesMath
Answer: -4x^3 + 8x^2 -5x + 12
Discussion:
Standard for for a polynomial meas tne power of x descend so standard fro of
8x^2 - 4x^3 +12 -5x is
-4x^3 + 8x^2 -5x + 12
Note: the problem statements shows this ":8x^2". I didn't know how to deal with the ":" character so simply ignored it.
Thank you,
Mr. B
Answer:
There should be 35 student in her class altogether.