The question might have some mistake since there are 2 multiplier of t. I found a similar question as follows:
The population P(t) of a culture of bacteria is given by P(t) = –1710t^2+ 92,000t + 10,000, where t is the time in hours since the culture was started. Determine the time at which the population is at a maximum. Round to the nearest hour.
Answer:
27 hours
Step-by-step explanation:
Equation of population P(t) = –1710t^2+ 92,000t + 10,000
Find the derivative of the function to find the critical value
dP/dt = -2(1710)t + 92000
= -3420t + 92000
Find the critical value by equating dP/dt = 0
-3420t + 92000 = 0
92000 = 3420t
t = 92000/3420 = 26.90
Check if it really have max value through 2nd derivative
d(dP)/dt^2 = -3420
2nd derivative is negative, hence it has maximum value
So, the time when it is maximum is 26.9 or 27 hours
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
2x-y=-8
+
y=2
2x=-6
x=-3
2*-3-y=-8
-6-y=-8
-y=-2
x=-3 and y=2
Answer:
hello
Step-by-step explanation:
