Answer:
At 25 = 6.8612mm
At 50 years = 5.422mm
Step-by-step explanation:
Equation,
d = 2.115Logₑa + 13.669
d = diameter of the pupil
a = number of years
Note : Logₑa = In a (check logarithmic rule)
d = 2.115Ina + 13.669
1. At 25 years,
d = -2.115In25 + 13.669
d = -2.115 × 3.2188 + 13.669
d = -6.807762 + 13.669
d = 6.8612mm
At 25 years, the pupil shrinks by 6.86mm
2. At 50 years,
d = -2.1158In50 + 13.669
d = -2.1158 * 3.912 + 13.669
d = -8.2770 + 13.699
d = 5.422mm
At 50 years, the pupil shirks by 5.422mm
To save this question, I had to plug in the values into the equation.
Solving for Logₑa might be difficult, so instead I used Inx which is the same thing. Afterwards, i substituted in the values and solve the equation for each years.
9514 1404 393
Answer:
- relative minimum -6√3 at x = -√3
- relative maximum 6√3 at x = √3
- decreasing on x < -√3 and x > √3
- increasing on -√3 < x < √3
- see below for a graph
Step-by-step explanation:
I find it convenient to draw the graph first when looking for relative extrema.
The function can be differentiated to get ...
f'(x) = -3x^2 +9
This is zero when ...
-3x^2 +9 = 0
x^2 = 3
x = ±√3 . . . . . x-values of relative extrema
Then the extreme values are ...
f(±√3) = x(9 -x^2) = (±√3)(9 -3) = ±6√3
The lower extreme (minimum) corresponds to the lower value of x (-√3), so the extrema are ...
(x, y) = (-√3, -6√3) and (√3, 6√3)
__
Since the leading coefficient is negative and the degree is odd, the function is decreasing for values of x below the minimum and above the maximum. It is increasing for values of x between the minimum and the maximum.
decreasing: x < -√3, and √3 < x
increasing: -√3 < x < √3
This can be modeled as:
P(y)=6X10^9(1.014)^(y-2000)
P(2010)=6,894,944,907
P(2075)=17,021,501,805
P(2120)=31,820,415,782 (all to the nearest whole person :) )
Answer:
you got this u are really smart
Step-by-step explanation:
016 is the correct answer