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harkovskaia [24]
3 years ago
13

with Judy jogging north and Jackie jogging west (on straight roads). When Jackie is 1 mile farther fro m the intersection than J

udy, the distance between them is 2 miles more than Judy’s distance from the intersection. How far is Jackie from the intersection?
Mathematics
1 answer:
miskamm [114]3 years ago
4 0
Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.

We convert the given information to equations, step by step.

Point 1: When Jackie is 1 mile farther from the intersection than Judy.

This means y is 1 mile more than x.

So,

y = 1 + x

Point 2: The distance between them is 2 miles more than Judy’s distance from the intersection. 

Distance between is x+ y.

So, x+y is 2 miles more than y.

x+y = y + 2
⇒
x = 2

From point 1 we have:

y = 1 + x = 1+ 2 = 3

So,

Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.
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Answer:

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Step-by-step explanation:

Let the number of years passed since 2010 to reach population more than 7000000 be 'x'.

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A population growth is an exponential growth and is modeled by the following function:

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Taking log on both sides, we get:

\log(P)=\log(P_0(1+r)^x)\\\log P=\log P_0+x\log (1+r)\\x\log (1+r)=\log P-\log P_0\\x\log(1+r)=\log(\frac{P}{P_0})\\x=\frac{\log(\frac{P}{P_0})}{\log(1+r)}

Plug in all the given values and solve for 'x'.

x=\frac{\log(\frac{700,000}{450,000})}{\log(1+0.05)}\\x=\frac{0.192}{0.021}=9.13\approx 10

So, for x > 9.13, the population is over 700,000. Therefore, from the tenth year after 2010, the population will be over 700,000.

Therefore, the tenth year after 2010 is 2020.

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