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tigry1 [53]
3 years ago
9

23. Assuming it is possible to walk between locations in a straight line, how much longer is it to walk from the museum to the z

oo than to walk from the museum to the park?​

Mathematics
1 answer:
kkurt [141]3 years ago
4 0

The distance from Museum to Zoo is 0.74 units longer than the distance from Museum to park.

Assuming straight distance between the points :

The distance formula between two points is :

d = \sqrt {\left( {x_2 - x_1 } \right)^2 + \left( {y_2 - y_1 } \right)^2 }

Museum coordinate (-2, - 3)

Zoo coordinate (3, 3)

Park coordinate (-3, 4)

Distance from Museum to Zoo:

d = \sqrt {\left( {3 - ( - 2)} \right)^2 + \left( {3 - ( - 3) } \right)^2 }

d \:  =  \sqrt{25 + 36}  =  \sqrt{61}  = 7.81

Distance from Museum to park :

d = \sqrt {\left( { - 3 - ( - 2)} \right)^2 + \left( {4 - ( - 3) } \right)^2 }

d \:  =  \sqrt{ 1 + 49}  =  \sqrt{50}  = 7.07

7.81 - 7.07 = 0.74

This means that :

Distance from Museum to Zoo is 0.74 units more than the distance from Museum to park.

Learn more : brainly.com/question/8283882

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Answer:

a)

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b)

The function is y(t) = 141200(0.89)^t

The number of bats in 2003 was of 13,729.

Step-by-step explanation:

Exponential decay function:

An exponential decay function has the following format:

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In which y(0) is the initial value, 1 - r is the decay factor and r is the decay rate, as a decimal.

A. Identify the initial amount, the decay factor, and the decay rate

In 1983 there were 141,200 bats living in the caves, which means that the initial amount is 141,200.

That number decreased by about 11% annually until 2003 means that the decay factor is 1 - 0.11 = 0.89 and the decay rate is 0.11 = 11%.

B. Write a function that models and number of that since 1983 then find the number of bats in 2003

With the data found in question A, we have that:

y(t) = y(0)(1-r)^t

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y(t) = 141200(0.89)^t

The function is y(t) = 141200(0.89)^t

Number in 2003

2003 is 2003 - 1983 = 20 years after 1983, so this is y(20).

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