So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
Graph -5 on the number line then use your slope (-1/2) and from -5 go one down and 2 right
lol it's so hard I can't proses
Answer:
<u>No, it's not a function.</u>
Step-by-step explanation:
This is not a function because the x-inputs repeat. It would be a function if each x-input was a different number, but the inputs 3 and 4 repeat twice. <u>Therefore, this is not a function.</u>
Answer:
24:8 which is also just 6:2 or 3:1
Step-by-step explanation:
There are many ways you can write this, and this is what I think hope this helps