H(x) = x2 + 1 k(x) = x – 2 (h + k)(2) = (h – k)(3) = evaluate 3h(2) + 2k(3) =

For this case we have the following functions:

We must perform each of the operations shown.

We have then:

For (h + k) (2):

Evaluating for x = 2 we have:

For (h - k) (3):

Evaluating for x = 3 we have:

For 3h (2) + 2k (3):

Value of function at x=a can be found by putting the value of x in the function. The value of (h+k)(2), (h-k)(3), and 3h(2)+2k(3) is 5, 6, and 17, respectively.

Given:

The given functions are:

Now, it is required to find the value of (h+k)(2), (h-k)(3), and 3h(2)+2k(3).

Solve for (h+k)(x) as,

So, the value of (h+k)(2) will be,

Similarly, the value of (h-k)(3) and 3h(2)+2k(3) will be,

Therefore,  the value of (h+k)(2), (h-k)(3), and 3h(2)+2k(3) is 5, 6, and 17, respectively.

For more details, refer to the link:

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step-by-step explanation:

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6