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.
Factor the following:
10 y^2 - 35 y + 30
Factor 5 out of 10 y^2 - 35 y + 30:
5 (2 y^2 - 7 y + 6)
Factor the quadratic 2 y^2 - 7 y + 6.
The coefficient of y^2 is 2 and the constant term is 6.
The product of 2 and 6 is 12.
The factors of 12 which sum to -7 are -3 and -4. So 2 y^2 - 7 y + 6 = 2 y^2 - 4 y - 3 y + 6 = y (2 y - 3) - 2 (2 y - 3):
5 y (2 y - 3) - 2 (2 y - 3)
Factor 2 y - 3 from y (2 y - 3) - 2 (2 y - 3):
Answer: 5 (2 y - 3) (y - 2)
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