<img src="https://tex.z-dn.net/?f=x%20%5C%3A%20varies%20%5C%3A%20as%20%5C%3A%20product%20%5C%3A%20u%20%5C%3A%20v%20%5C%3A%20and%

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3 years ago

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20%5C%3A%20inversely%20%5C%5C%20as%20%5C%3A%20their%20%5C%3A%20sum%20%5C%3A%20if%20%5C%3A%20x%20-%203%20%5C%3A%20when%20%5C%3A%20u%20%3D%203%20%5C%3A%20%20%5C%5C%20and%20%5C%3A%20v%20%3D%201%20%5C%3A%20what%20%5C%3A%20is%20%5C%3A%20the%20%5C%3A%20value%20%5C%3A%20of%20%5C%3A%20x%20%5C%3A%20if%20%5C%5C%20u%20%3D%203%20%5C%3A%20and%20%5C%3A%20v%20%3D%203" id="TexFormula1" title="x \: varies \: as \: product \: u \: v \: and \: inversely \\ as \: their \: sum \: if \: x - 3 \: when \: u = 3 \: \\ and \: v = 1 \: what \: is \: the \: value \: of \: x \: if \\ u = 3 \: and \: v = 3" alt="x \: varies \: as \: product \: u \: v \: and \: inversely \\ as \: their \: sum \: if \: x - 3 \: when \: u = 3 \: \\ and \: v = 1 \: what \: is \: the \: value \: of \: x \: if \\ u = 3 \: and \: v = 3" align="absmiddle" class="latex-formula">
Pls help me with this

1 answer:

EleoNora [17]3 years ago

4 0

[Note that you don't have to evaluate k to solve the problem]

$x = \frac{kuv}{(u+v)}$

That means that

$x \frac{(u+v)}{(uv)} = k, a \: constant$

So, you want $x$ such that

$x \frac{(3+3)}{(3 \times 3)} = 3 \frac{(3+1)}{(3 \times 1)}$

$x=6$

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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

DiKsa [7]

Answer:

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:

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

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

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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