Alan

At t = 0, A(0) = 550 years

Half of this occurs when  550/2 = 550*e^(-0.158t)

1/2 = e^(-0.158t)

ln(1/2) = -0.158t

-ln(2) = -0.158t

t = ln(2)/0.158  years  = 4.387 years ot t = 4.39 years to the nearest hundredth of a year.

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An Internet search suggests that in 1936 the black Rapids glacier moved an average of 115 ft a day for a total of 4 miles.

1 ft is 0.0254*12 metres

1 day is 3600*24 seconds

So 115 ft per day is a speed of 115*0.0254*12/(3600*24) m/s

I'll leave you to crunch the numbers.

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300:(23-3) → 300:20 → 30:2 ⇒ 15:1

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You need to supply the distance the metal stakes moved and the time they took.

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This factors as (a - 3)^2 = 0 so there is a double root at a = 3

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This factors as (x + 1)^2 = 0 so there is a double root at x = -1

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z= kxy/t^2

5 = k*2*(-3)/(1/4).    Hence k = -5/24

z = -(5/24)xy/t^2

When x = -12/(5z)^2 then z = -(5/24)(-12)y/(5zt)^2 or z^3 = y/(10t^2)

With z = 2/5 we have 8/125 = y/(10t^2) so that t = (5/4)y^(1/2)

Hence if f(y) = (5/4)y^(1/2) then f(y) = t when the above conditions hold.

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You can see immediately that neither a nor d can be correct as they give negative heights for all values of t, whereas the question specifies positive heights.

You can also rule out b because this could only give a height of 161 at time t = 0, and for greater times, including t= 1, the height would be less than 161, which would conflict with the information given in the question.

Substitute one of the given times into c and e to see which produces the corresponding specified height.

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(3/5)total = 12

total = 12*5/3

total = 20

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Try WolframAlpha

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