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2 years ago

The graph below shows the distance Sean rode his bike. Does this represent a proportional relationship?

Mathematics

2 answers:

dybincka [34]2 years ago

7 0

Answer:

Step-by-step explanation:

First the line is a straight line. It starts at zero and increases 10 miles every hour. You cannot find a better example of y = kx

I don't know what A means, but it is not the answer. It does not represent what is in the graph.

The information from the graph is just fine. B is not the answer.

D is incorrect. The graph says he rode 10 miles in each hour.

The answer is C. For every hour on his bike, he rides 10 miles. That is a proportional relationship, especially when he starts at 0.

Black_prince [1.1K]2 years ago

4 0

We can see that for every 1 hour he increases 10 miles in his distance, so we can say it's constant, he started at 0 and increased 10 miles per hour.

Alternative C.

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

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1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

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2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

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