Answer:
The town's annual energy consumption will be of 6.57 trillons of BTU after 9 years.
Step-by-step explanation:
The annual energy consumption of the town where Camilla lives increases at a rate that is proportional at any time to the energy consumption at that time.
This means that the consumption after t years is given by the following differential equation:
In which k is the growth rate.
The solution is, applying separation of variables:
In which C(0) is the initial consumption.
The town consumed 4.4 trillion British thermal units (BTUs) initially.
This means that 
So
5.5 trillion BTUs annually after 5 years.
This means that 
. We use this to find k. So
So
After 9 years?
This is C(9). So
The town's annual energy consumption will be of 6.57 trillons of BTU after 9 years.
Answer:
3.85 hours
Step-by-step explanation:
We have that the model equation in this case would be of the following type, being "and" the concentration of bacteria:
y = a * e ^ (b * t)
where a and b are constants and t is time.
We know that when the time is 0, we know that there are 100,000 bacteria, therefore:
100000 = a * e ^ (b * 0)
100000 = a * 1
a = 100000
they tell us that when the time is 2 hours, the amount doubles, that is:
200000 = a * e ^ (b * 2)
already knowing that a equals 100,000
e ^ (b * 2) = 2
b * 2 = ln 2
b = (ln 2) / 2
b = 0.3465
Having the value of the constants, we will calculate the value of the time when there are 380000, that is:
380000 = 100000 * (e ^ 0.3465 * t)
3.8 = e ^ 0.3465 * t
ln 3.8 = 0.3465 * t
t = 1.335 / 0.3465
t = 3.85
That is to say that in order to reach this concentration 3.85 hours must pass
3.10/155= 0.02 } subtract = 0.07 more
7.65/85= 0.09 }
i feel like u are going to delete this but if this helped please don't delete it the answer is, Simon is correct because even though the input values are opposite in the reflected function, any real number can be an input.
Answer:
Step-by-step explanation:
The given quadratic equation is
2x^2+3x-8 = 0
To find the roots of the equation. We will apply the general formula for quadratic equations
x = -b ± √b^2 - 4ac]/2a
from the equation,
a = 2
b = 3
c = -8
It becomes
x = [- 3 ± √3^2 - 4(2 × -8)]/2×2
x = - 3 ± √9 - 4(- 16)]/2×2
x = [- 3 ± √9 + 64]/2×2
x = [- 3 ± √73]/4
x = [- 3 ± 8.544]/4
x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4
x = 5.544/4 or - 11.544/4
x = 1.386 or x = - 2.886
The positive solution is 1.39 rounded up to the nearest hundredth
