Answer:

Step-by-step explanation:
We are given:

![interval = [a,b] = [0,2]](https://tex.z-dn.net/?f=interval%20%3D%20%5Ba%2Cb%5D%20%3D%20%5B0%2C2%5D)
Since
⇒ 
Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.

Note:
If it will be asked to find right endpoint too,

The average of left and right endpoint Riemann sums will give approximate result of the area under
and it can be compared with the result of integral of the same function in the interval given.
So, 

Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.
Answer:
There are many. Two examples are

Step-by-step explanation:
There are many examples. The simplest is
1 -

It is trivial that

2 -

That function is injective as well.

An example of a function that is NOT injective is

Notice that

For this case we have that by definition, the domain of a function, is given for all the values for which the function is defined.
We have:

The given function is not defined when the denominator is equal to zero. That is to say:

To find the roots we factor, we look for two numbers that when multiplied give as a result "8" and when added as a result "-6". These numbers are:

Thus, the factored polynomial is:

That is to say:

Makes the denominator of the function 0.
Then the domain is given by:
All real numbers, except 2 and 4.
Answer:
x |x≠2,4
C, i think you forgot an x
(x+11)+(x-7)
x+11+x-7
x+x+4
2x+4
Answer:
18 apples equals 6 bananas