Answer:
Step-by-step explanation:
We are given that,
The fall of the acorn is represented by the function,
Equating S(t) to 0 gives,
i.e. 
i.e. 
i.e. 
Since, time cannot be negative.
So, 
i.e. 1.25 seconds is the time when the acorn will reach the ground.
Thus, from the figure below, we see that,
The interval in which the acorn is in the moving air is .
300,000,000+9,000,000+90,000+9,000+900+90
Answer:
This would be shifted down 8 and made 3 times less steep.
Step-by-step explanation:
In order to determine these transformations, we first need to compare the constants at the end. This will determine the up or downward shift. Since the f(x) is 5 and the g(x) is -3, we know that it went down 8.
Next we compare the coefficients of x. Since the f(x) is 6 and the g(x) is 2, we know that it is 3 times less steep.
Answer:
A.
A(n) = P(1 + i)^n-1, where n is a positive integer
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.
