the instantaneous growth rate r of a colony of bacteria t hours after the start of an experiment is given by the function r=0.01t^3-0.07t^2+0.07t+0.15 for 0≤t≤7. find the times for which the instantaneous growth rate is zero.

t = 5,-1,3

Step-by-step explanation:

r=0.01t^3-0.07t^2+0.07t+0.15

For simplification let's multiply the equation by 100

100r = t³ - 7t² + 7t + 15

When r= 0

t³ - 7t² + 7t + 15= 0

Let's look for the value of t.

Let's try a possible division

(t³ - 7t² + 7t + 15)/(t-5) = t² -2t -3

So we've gotten one as t-5

Let's factorize t² -2t -3

= t² +t -3t -3

= t(t+1) -3(t+1)

= (t+1)(t-3)

So we have

(t-5)(t+1)(t-3)

What it means is that the possible values are when

t = 5,-1,3

These are also called the roots of the equation

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

the answer is 13/48

step-by-step explanation: