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3 years ago

10

I need help asap with math

1 answer:

Dmitrij [34]3 years ago

4 0

I'm not even going to bother with the substitution method.
Instead of all that, I just want to show you something:

Take the FIRST equation:

y - 2x = 8
Add 2x to
each side: y = 8 + 2x

Multiply each
side by 2 : 2y = 16 + 4x

Look at that ! The first equation says the same thing as the second one !
Both of them are actually the same equation.
You actually have only one equation, with 'x' and 'y' in it.
If you graph it, the graph is a straight line, and EVERY point
on the line is a solution of the equation.

==> There are an infinite number of solutions, because
they only gave you one equation, with two unknown variables.
To find a unique solution, you'd need two equations.

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4 years ago

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3 years ago

The volume V of a gas varies inversely as the pressure P and directly as the temperature A certain gas has a volume of 10 liters

Ad libitum [116K]

Answer:

Joint variation states that:

If z is directly proportional to x and inversely proportional to y,

i.e, $z \propto x$ , $z \propto \frac{1}{y}$

then the equation is in the form: $z = k\frac{x}{y}$ where k is the constant of variation.

From the given information: The volume V of a gas varies inversely as the pressure P and directly as the temperature

i.,e $V \propto T$ , $V \propto \frac{1}{P}$

by definition of joint variation:

$V = k\frac{T}{P}$ .....[1] where k is the constant of variation.

It is also given that a certain gas has a volume of 10 liters(L) , a temperature of 300 kelvins(K), and a pressure of 1.5 atmosphere(atm).

Substitute these given values in [1] to solve for k;

$10 = k\frac{300}{1.5}$

Simplify:

$10 = 200k$

Divide by 200 both sides we have;

$k = \frac{10}{200} =\frac{1}{20}$

Now, if a gas has a temperature of 400 kelvins and a pressure of 5 atm.

To find the volume (V);

Now substitute the given data and value of k in [1] we have;

$V = \frac{1}{20} \cdot \frac{400}{5} = \frac{20}{5} = 4$

Therefore, the Volume of a gas is, 4 liters

5 0

3 years ago