An=-31+(n-1)-11 find the eleventh term
a11 = -31 + (-11) (11-1)
a11 = -31 +-11(10)
a11 =-31 +-110
a11 = -142
I'm tentatively changing my answer to say this kind of relies on practical knowledge of how stores tend to operate. If 20 coupons are given out, the store has sold all 500 shirts, arguably at a loss to the retailer. They have to have more shirts in stock to be sold at full price because, well, that's how they make money. It's more likely that such a store would carry more than just 500 shirts at the start of each day, so A is (probably) wrong.
Answer:
Step-by-step explanation:
This means that 4/5 repeats itself 3 times.
So,
![\sf = \frac{4}{5} * \frac{4}{5} * \frac{4}{5} \\\\= \frac{64}{125} \\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20%3D%20%5Cfrac%7B4%7D%7B5%7D%20%2A%20%5Cfrac%7B4%7D%7B5%7D%20%2A%20%5Cfrac%7B4%7D%7B5%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B64%7D%7B125%7D%20%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Answer:
The speed of the first train is 45 mph and the speed of the second train is 75 mph
Step-by-step explanation:
Let x represent the speed of the first train in mph. Since the second train, is 30 mph faster then the first, therefore the speed of the second train is (x + 30).
The first train leaves at 1:00 pm, therefore at 6:00 pm, the time taken is 5 hours. Therefore the distance covered by the first train at 6:00 pm = x mph * 5 hours = 5x miles
The second train leaves at 3:00 pm, therefore at 6:00 pm, the time taken is 3 hours. Therefore the distance covered by the second train at 6:00 pm = (x + 30) mph * 3 hours = (3x + 90) miles
Since the second train overtakes the first at 6:00 pm, hence:
3x + 90 = 5x
2x = 90
x = 45
Therefore the speed of the first train is 45 mph and the speed of the second train is 75 mph (45 mph + 30 mph).
a. The difference between two outputs that are 1 unit apart.
You need to Use y2 - y1 / x2 - x1 to find the difference
I will choose x2 as 1 and x1 as 0
(29 - 21) / (1 - 0)
8/1 so The difference is 8 per 1 unit.
b. Use the same formula
I will choose -3 as x2 and -5 as x1
(5 - (-11)) / (-3 - (-5))
(5 + 11) / (-3 + 5)
16 / 2 so the difference is 16 per 2 units.
c. I will choose 2 as x2 and -1 as x1
(45 - 21) / (2 - (-1))
24/3 so the difference is 24 per 3 units.
d. The ratios of the differences to the input intervals reduced all equal each-other, which is 8 per 1 unit.
