4,498,250,098
Hope that helped:D
Let's call this line y=mx+C, whereby 'm' will be its gradient and 'C' will be its constant.
If this line is parallel to the line you've just mentioned, it will have a gradient 2/3. We know this, because when we re-arrange the equation you've given us, we get...

So, at the moment, our parallel line looks like this...
y=(2/3)*x + C
However, you mentioned that this line passes through the point Q(1, -2). If this is the case, for the line (almost complete) above, when x=1, y=-2. With this information, we can figure out the constant of the line we want to find.
-2=(2/3)*(1) + C
Therefore:
C = - 2 - (2/3)
C = - 6/3 - 2/3
C = - 8/3
This means that the line you are looking for is:
y=(2/3)*x - (8/3)
Let's find out if this is truly the case with a handy graphing app... Well, it turns out that I'm correct.
Answer:
((2 x^2 + 1)^2)/(x^2)
Step-by-step explanation:
Simplify the following:
(2 x + 1/x)^2
Hint: | Put the fractions in 2 x + 1/x over a common denominator.
Put each term in 2 x + 1/x over the common denominator x: 2 x + 1/x = (2 x^2)/x + 1/x:
((2 x^2)/x + 1/x)^2
Hint: | Combine (2 x^2)/x + 1/x into a single fraction.
(2 x^2)/x + 1/x = (2 x^2 + 1)/x:
((2 x^2 + 1)/x)^2
Hint: | Distribute exponents over quotients in ((2 x^2 + 1)/x)^2.
Multiply each exponent in (2 x^2 + 1)/x by 2:
Answer: ((2 x^2 + 1)^2)/(x^2)
Answer: $20
$48/24 = $2 for each candy bar
10 candy bars * $2 = $20