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2 years ago

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Your bedroom has a rectangular shape and you want to measure its size. You use a tape that is precise to 0.001 mm and find that

the shortest wall in the room is 3.547 mm long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 mm. You measure the second wall to be 4.79. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
a. 30 m^2
b. 31.4 m^2
c. 31.37 m^2
d. 31.371 m^2

1 answer:

avanturin [10]2 years ago

4 0

This question is in two parts. This is not the correct multiple choice options for this part a.

The second part had the option

b)If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?

a)30 m^2

b) 31.4 m^2

c)31.37 m^2

d)31.371 m^2

Answer:

A. 17.0 m²

B. 31.4 m²

Explanation:

The formula for the calculation of the area of a rectangle is given as

Area = length x width

The length = 3.547 m

The width = 4.79 m

Then area = 3.547 x 4.79

= 16.990m²

When approximated = 17.0m²

This is the most precise measurement for the area of the bedroom.

B.

We solve b using this formula

Area = pi(diameter/2)^2

= 3.14(6.32/2)²

= 3.14 x 9.9856

= 31.4 m²

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Explanation:

Part a

For this case we have the following differential equation:

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With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

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Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

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Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

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Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

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