The summand (R?) is missing, but we can always come up with another one.
Divide the interval [0, 1] into 
subintervals of equal length 
:
![[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]](https://tex.z-dn.net/?f=%5B0%2C1%5D%3D%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B1-%5Cdfrac1n%2C1%5Cright%5D)
Let's consider a left-endpoint sum, so that we take values of 
where 
is given by the sequence
with 
. Then the definite integral is equal to the Riemann sum
19
If she has already watched 1.5 hours and she can only watch 5 hours per week, then we have the inequality:
1.5 + T ≤ 5
where T is is the number of hours she can watch more before reaching her limit
20
If he needs 100 points, already has 40 and earns 2 for every movie, we have the inequality:
40 + 2m ≥ 100
where m is the number of movies he needs to watch
21
We can have the equation:
x + 1 = x + 2 - 1
This would result to:
x + 1 = x + 1
which have the same terms on both sides.
Further, the equation would be:
0 = 0
Answer: B is correct
Step-by-step explanation:
if you do 2 ÷ 5 you get 0.4. But that isn't an option. So, the only other possibility would be -2 ÷ 5 which gives us -0.4, and that makes sense.
I'm not the greatest at explaining this one but hopefully, this helps!
