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3 years ago

Can you please help me with this problem?

Mathematics

1 answer:

Hoochie [10]3 years ago

6 0

Here you have to find which each variable is, for this you start of picking one equation,

x + 2y + 6z = 4
-3x + 2y - 2 = -4
4x + 2z = 16

depending the equation you pick you multiply that by a certain number that will give you the opposite of one of the other equations,
-1(x + 2y + 6z = 4)
= -x -2y - 6z = -4

With this you add or subtract it with the equation that has the same number or variable, or both,
In this case it will be the equation,
-3x + 2y + 6z = 4
You can use this one or the third equation since both have a positive 2y which will cancel with -2y from the new equation,
-x - 2y - 6z = -4
-3x + 2y -z = -4

= -4x -7z = -8
Now you since you just eliminated the variable (y) you now have 2 variables, and the last equation has only 2 variables, meaning now you find the answer to those to equations,

-4x -7z = -8
4x + 2z = 16

= -5z = 8
Now leave the variable by itself,
z = 8/5
Now you found the variable (z), with this just substitute on one of the equations we used to find (z) so you can find (x), after that substitute those answered to on of the original equations so you can find (y)

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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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4 years ago

Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!!

Naddik [55]

X=7
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aleksandrvk [35]

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