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>
Lol this one is kind of confusing but i think it’s C because there can be a centipede that is missing a leg lol
Answer:
Let the ten's digit no. be x and one's digit no. be y.
So the no. will be = 10x+y.
Given : x+y=9-----(I)
9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)
On solving I and II simultaneously you will get x=1 and y=8.
Therefore your desired no. is 18.
Step-by-step explanation:
Hope it is helpful.....
Answer:
The solution would be C, 28/9
