Alake polluted by bacteria is treated with an antibacterial chemical. aftertdays, thenumber n of bacteria per milliliter of water is approximated byn(t) = 20(t12−ln(t12))+ 30for 1≤t≤15.a. when will the number of bacteria be a minimum? b. what is the minimum number of bacteria during this time? c. when will the number of bacteria be a maximum? d. what is the maximum number of bacteria during this time?

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

Extrema values of functions

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

(a)

First, we take its derivative

Solve N'(t)=0

Simplifying

Solving for t

Only t=1 belongs to the valid interval

Taking the second derivative

Which is always positive, so t=1 is a minimum

(b)

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=4391.7 is a maximum

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x