Question

A common factor of 5p 2q 2 and 25pq​

Answer 1

Answer:

Function f(t) is given as:

\boxed{ \bf{f(t) = 500.e^{0.05\:t}}}

f(t)=500.e

0.05t

here, 't' is the time and is measured in minutes.

We are required to find the population of bacteria after 4 hours.

Hence in 4 hours, we have 4 × 60 = 240 minutes. Hence the value of 't' is 240 minutes.

Substituting the values we get:

\begin{gathered}f(240) = 500.e^{(0.05 \times 240)}\\\\f(240) = 500 \times e^{(12)}\\\\f(240) = 500 \times 1.62 \times 10^5\\\\\boxed{ \bf{f(240) = 8.13 \times 10^7}}\end{gathered}

f(240)=500.e

(0.05×240)

f(240)=500×e

(12)

f(240)=500×1.62×10

5

f(240)=8.13×10

7

Hence the amount of bacteria present after 4 hours is 8.13 × 10⁷ (approx

Answer 2

Answer:

1

Step-by-step explanation:

they have no common factor so 1 is the only fator for all number...