Answer:
Step-by-step explanation:
We want to find the Riemann sum for 
with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:
where 
.
Step 1: Find 
We have that 
Therefore, 
Step 2: Divide the interval ![\left[0,\frac{3 \pi}{4}\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D)
into n = 6 sub-intervals of length 
![a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b](https://tex.z-dn.net/?f=a%3D%5Cleft%5B0%2C%20%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D%3Db)
Step 3: Evaluate the function at the left endpoints
Step 4: Apply the Left Riemann Sum formula
Answer:
4 : 9
Step-by-step explanation:
The requested ratio is ...
tile length : board length = (2/3 ft) : (1/2 yd) = (8 in) : (18 in)
= 8 : 18 = 4 : 9
_____
To make a unitless ratio, both parts must have the same units. Here, we chose to express the lengths in inches. We could have used feet or yards as well. (2/3 ft)×(1 yard)/(3 ft) = 2/9 yd or (1/2 yd)(3 ft/yd) = 3/2 ft. You get the same ratio with any of these:
(2/9 yd) : (1/2 yd) = (4 yd) : (9 yd) = 4 : 9 . . . . . multiply by 18
(2/3 ft) : (3/2 ft) = (4 ft) : (9 ft) = 4 : 9 . . . . . multiply by 6
P= 2(b+2) + 2(b)- it would be 2 (width) plus 2( length)
