The average of six numbers is 7. If two of the six numbers are removed, the average of the remaining numbers is 8. What is the sum of the two numbers which were removed?

10

Step-by-step explanation:

Mean = sum of the terms / number of the terms

Let's assume that the sum of the terms is x

So it'd be: x/6 = 7  

x = 6 X 7 = 42

Since two numbers are removed, there are only 4 terms

So it'd be: x/4 = 8

x = 8 X 4 = 32

Finally work out the difference between the two averages:

42 - 32 = 10

So the sum of the 2 numbers was 10.

Hope this helps :)

F( x) = 1/(x^ 2 - 3x + 2)

Given function should be maximum when (x^ 2 - 3x + 2) will be minimum.

Let , y = (x^ 2 - 3x + 2)

Differentiate with respect to x, we get

dy/dx = 2x - 3

For maximum or minimum we will put dy/dx = 0

=> x= 3/2

Now, we find the second derivative.

=> d^2y/dx^2 = 2 (+ ve)

That means (x^ 2 - 3x + 2) is minimum.

Therefore f(x) = 1/(x^ 2 - 3x + 2) is maximum.

=> Maximum value at 3/2 is

=> 1/(9/4 - 9/2 + 2)

=> 4/(9 - 18 + 8)

=> - 4

Hence, the maximum value of the function is - 4