The answer is B) 2 and 3
When you do a Riemann's sum, you split the interval into columns. In choice 3 we are told that there are 4 equal subdivisions meaning that each of the 4 columns are 1 wide. Since it is a left Riemann's sum we add the height of the column where the LEFT side crosses the curve. For [-1, 0], the height is -1, for [0, 1] it is 0, for [1, 2] it is 2, and for [2, 3] it is 3. Now add the heights together to get 4. Then you multiply by the width, but since the columns are 1 wide, the answer is still just 4.
Everybody solves things differently. For me, I find that by writing everything down in my agenda, it helps me organize my thoughts and I'm able to visually see what assignment(s) are due soon.
Since your history exam is coming up, I would highly suggest you to study a bit each day. It's a bit of a stretch since May isn't even here yet, but it's better to start early.
Once again, although May isn't here yet, I would suggest working on whatever is due in that month. Believe me, doing it last minute is seriously so stressful and you'll regret it. (I'm saying this from experience.)
Just remember to give yourself little breaks and be sure to have enough sleep! It's fine if there will be times where you find yourself staying up the whole night just to get your work done. But, don't do it often as that is not good for your health.
Some tips for when you find yourself getting stressed out while working. I heard that tea, specifically Chamomile tea, helps reduce stress. Perhaps listening to music or taking a walk outside will help.
I hope my tips help you at least a bit.
The first thing that she opens is the door to let them in aha
Answer:
Length of arc = 21.99 inch (Approx.)
Explanation:
Given:
Radius of given circle = 6 inch
Central angle = 210°
Find:
Length of arc
Computation:
Length of arc = [θ/360][2πr]
Length of arc = [210/360][2(22/7)(6)]
Length of arc = [0.583][12(3.1428)]
Length of arc = [0.583][12(3.1428)]
Length of arc = [0.583][37.7136]
Length of arc = 21.987
Length of arc = 21.99 inch (Approx.)
