The answer is 6.89 because
6.89+14.52=21.41
So 21.41+(-14.52)=6.89
I'll let <em>h</em> = <em>ax</em>, so the limit is
![\displaystyle\lim_{h\to0}\frac{(x+h)^2-2(x+h)+1-(x^2-2x+1)}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%28x%2Bh%29%5E2-2%28x%2Bh%29%2B1-%28x%5E2-2x%2B1%29%7Dh)
i.e. the derivative of
.
Expand the numerator to see several terms that get eliminated:
![(x+h)^2-2(x+h)+1-(x^2-2x+1)=x^2+2xh+h^2-2x-2h+1-x^2+2x-1=2xh+h^2-2h](https://tex.z-dn.net/?f=%28x%2Bh%29%5E2-2%28x%2Bh%29%2B1-%28x%5E2-2x%2B1%29%3Dx%5E2%2B2xh%2Bh%5E2-2x-2h%2B1-x%5E2%2B2x-1%3D2xh%2Bh%5E2-2h)
So we have
![\displaystyle\lim_{h\to0}\frac{2xh+h^2-2h}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B2xh%2Bh%5E2-2h%7Dh)
Since <em>h</em> ≠ 0 (because it is approaching 0 but never actually reaching 0), we can cancel the factor of <em>h</em> in both numerator and denominator, then plug in <em>h</em> = 0:
![\displaystyle\lim_{h\to0}(2x+h-2)=\boxed{2x-2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%282x%2Bh-2%29%3D%5Cboxed%7B2x-2%7D)
Answer:
Uhh, I think the answer is <em>b</em>?
Step-by-step explanation:
if you look at the area <em>d</em>, it's kind of the same as 18 mm but it isn't. It's a few inches off the same inch as 18 mm, so it MIGHT be <em>b</em>.
(I MIGHT BE WRONG, IF I AM, I AM SORRY. I'M NOT THAT SMART, SO EXPECT SOME WRONG ANSWERS OR NOT.)
Treat it like a fraction
18b^2/45b
divide top and bottom by b
18b/45
divide top and bottom by 3
6b/15
divide top and bottom by 3 again
2b/5
2b:5 is the simpliest form of the ratio
Write the set of points from -6 to 0 but excluding -4 and 0 as a union of intervals
First we take the interval -6 to 0. In that -4 and 0 are excluded.
So we split the interval -6 to 0.
Start with -6 and go up to -4. -4 is excluded so we break at -4. Also we use parenthesis for -4.
Interval becomes [-6,-4) . It says -6 included but -4 excluded.
Next interval starts at -4 and ends at 0. -4 and 0 are excluded so we use parenthesis not square brackets
(-4,0)
Now we take union of both intervals
[-6,-4) U (-4,0) --- Interval from -6 to 0 but excluding -4 and 0