Answer:
79, 81, 83, and 85
Step-by-step explanation:
x = 1st integer
x+2 = 2nd
x+4 = 3rd
x+6 = 4th
328 = sum
x+x+2+x+4+x+6 = 328
328/4 = 82
The 4 integers are the 2 odd integers on either side of 82 and closest to 82
That would be 79, 81, 83, and 85.
79 + 81 + 83 + 85 = 328
Answer:
yes -5 is greater than -10 because on a timeline -5 comes before -10
which means its greater
I believe
4 and 6
4/1 and 6/1<span />
<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
Starting from what I know (the difference between ribbons), I decided to go from the bottom and work my way up. I then noticed a pattern (each sum was three more of the previous one), and decided to keep my pattern of the three numbers but not have to do any major mental work and instead add three to the previous sum until I got to 38.
