The number of bacteria in a culture is growing at a rate of 3,000e^(2t/5) per unit of time t. At t=0, the number of bacteria present was 7,500. Find the number present at t=5.
Answer: 7,500e^2
X=9 because u divide 27 and 3
1) 52 + 33= 85
2) 26 + 42 = 68 or 26 - 42 = - 16
3) 72 - 2015 = =1943
4) 32 x 22 = 704
