For the function y = (x^3 − 3)(x^2 − 5x + 1) at (2, −25) find the following.
(a) the slope of the tangent line
(b) the instantaneous rate of change of the function

(a)

(b)

Step-by-step explanation:

We have been a function . We are asked to find the instantaneous rate of change of the function and the slope of the tangent line at point .

(a) First of all, we will find the derivative of our given function using product rule.

Now, we will substitute in our derivative function to find slope of tangent line as:

Therefore, the slope of the tangent line is -65 at point .

(b) We know that instantaneous rate of change of the function at a point is equal to the derivative of the function at that point.

We already figured it out that derivative of our given function at is , therefore, the instantaneous rate of change of the function is also at point .

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10