Answer:


Step-by-step explanation:
Given system of equations:

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

Factor the quadratic:

Apply the <u>zero-product property</u> and solve for x:


Substitute the found values of x into the <u>second equation</u> and solve for y:


Therefore, the solutions are:


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
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:
tens=5
Step-by-step explanation: