Splitting up [0, 3] into 
equally-spaced subintervals of length 
gives the partition
![\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac3n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac3n%2C%20%5Cdfrac6n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac6n%2C%20%5Cdfrac9n%5Cright%5D%20%5Ccup%20%5Ccdots%20%5Ccup%20%5Cleft%5B%5Cdfrac%7B3%28n-1%29%7Dn%2C%203%5Cright%5D)
where the right endpoint of the 
-th subinterval is given by the sequence
for 
.
Then the definite integral is given by the infinite Riemann sum
Answer:
3
Step-by-step explanation:
let the no of people who share the meal be x
each member paid $17
total cost of meal=$51
so ur equation is
$17x=$51
x=$51/$17(here $ and $ also gets cancel)
x=3
therefore three menber shared the meal.
Note:when in multiplication when u transfers the number to other side it changes into mltiplication.For example
$17=$51
here $17 is multiplied with x and when we bring $17 to right hand side it divides $51.
Answer:
rationalizing
Step-by-step explanation:
Answer:
ok
Step-by-step explanation:
what do u mean by the simplest form?
