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:

Step-by-step explanation:
Given:
Number of bushes planted = 5
Minimum number of bushes to be planted = 12
Let the number of bushes planted after planting 5 bushes be
.
Since, Tamara has already planted 5 bushes, total number of bushes planted is given as:

Now, as per question, total number of bushes should be 12 or greater than 12.
Therefore,
.
I think it’s the last one I am not sure tho
f(h(x))= 2x -21
Step-by-step explanation:
f(x)= x^3 - 6
h(x)=\sqrt[3]{2x-15}
WE need to find f(h(x)), use composition of functions
Plug in h(x)
f(h(x))=f(\sqrt[3]{2x-15})
Now we plug in f(x) in f(x)
f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6
cube and cube root will get cancelled
f(h(x))= 2x-15 -6= 2 x-21