So let f(x) = x^(2/3)
<span>Then let f'(x) = 2/3 x^(-1/3) = 2 / (3x^(1/3)) </span>
<span>When x = 8, </span>
<span>f(8) = 8^(2/3) = 4 </span>
<span>f'(8) = 2 / (3*8^(1/3)) = 1/3 </span>
<span>So near x = 8, the linear approximation is </span>
<span>f(x) ≈ f(8) + f'(8) (x - 8) </span>
<span>f(x) ≈ 4 + 1/3 (x - 8) </span>
<span>So the linear approximation for x = 8.03 is... </span>
<span>f(8.03) ≈ 4 + 1/3 (8.03 - 8) </span>
<span>f(8.03) ≈ 4 + 1/3 (0.03) </span>
<span>f(8.03) ≈ 4.01 </span>
<span>8.03^(2/3) ≈ 4.01 </span>
Answer:
D or 4 square root 3
Step-by-step explanation:
using distance formula -12 square rooted - 2 square root 3 = -4squareroot3. So, after that you would square giving 48 then square root giving 4 square root 3. (-4 square root 2 and -32 square rooted cancel out giving 0)
Answer:
2
Step-by-step explanation:
3/2 + (-k) + (-2) =
= 3/2 + [-(-5/2)] - 2
= 3/2 + 5/2 - 2
= 8/2 - 2
= 4 - 2
= 2
Move your cursor man it’s covering a critical value needed to answer
