Answer:
y=Ae^(1.25t)
Step-by-step explanation:
From the expression y=Ae^kt
After two days of the experiment, y = 49 million, t=2
After four days of the experiment, y= 600.25 million, t=4
A is the amount of bacteria present at time zero and t is the time after the experiment (in days)
At t=2 and y =49
49=Ae^2k…………….. (1)
At t=4 and y = 600.25
600.25=Ae^4k………… (2)
Divide equation (2) by equation (1)
600.25/49=(Ae^4k)/(Ae^2k )
12.25=e^2k
Take natural log of both sides
ln(12.25) =2k
2.505 =2k
k=1.25
The exponential equation that models this situation is y=Ae^(1.25t)
Complete the square since can't factor
2/2=1
add 1 to both sides
x²+2x+1-6=1
factor perfect square
(x+1)²-6=1
add 6 to both sides
(x+1)²=7
sqrt both sides
x+1=+/-√7
minus 1 both sides
x=-1+/-√7
so x=-1+√7 or x=-1-√7
Answer: the anser is
Step-by-step explanation:
9/24= 0.375 0.375x40=15
w=15 liters
Answer: P = $ 12000
r = 14%
t = 1 (for first year)
I = (P X r X t)/100
∴ I = (12000 X 14 X 1)/100
= 120 X 14
= $ 1680 <---------- (Interest on loan at the end of first year)
∴ Total amount owing at the end of first year = (P + I)
= (12000 + 1680)
= $ 13680
Repayment = $ 7800
Amount still outstanding (at the start of second year) = 13680 - 7800
= $ 5880
Interest on the outstanding amount at the end of second year,
P (new) = $ 5880
r (same) = 14%
t = 1 (for the current second year)
∴ I = (P X r X t)/100
= (5880 X 14 X 1)/100
= 82320 / 100
= $ 823.2 <-------------------------- (Interest on outstanding amount at the end of second year)
Answer:
f(4) = 11
Step-by-step explanation:
To evaluate f(4) substitute x = 4 into f(x)
f(4) = 3(4) - 1 = 12 - 1 = 11
