Answer: 
Step-by-step explanation:
Let w denotes a seven letter word which Sal will need to make to break the record.
Since you get 50 points for each seven-letter and 100 points for playing the game.
Then the total points earned by you by making w seven letters words will be:

Since, Sal wants to break that record, and needs 18,000 points or more to do so.
Then , the required inequality will be :

Answer:
25,000,000
Step-by-step explanation:
Hope this helps!!!
I think the answer is D. The other answers are unreasonable. Wyatt would not get the promotion and just because Tammy's lowest value is bigger than Wyatt's doesn't mean that Tammy gets the promotion. The only reasonable answer is D.
Answer:
Step-by-step explanation:
<u>Given system:</u>
<u>Multiply the first equation by 7 and add up the equations:</u>
- -7x + 7y + 7x + 3y = 7*8 - 16
- 10y = 40
- y = 4
<u>Find y:</u>
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

