TheXSquaredFactor

\(\pi*7^2=49\pi\approx153.93804003\)

In 12 days, Matthew will go hiking and swimming on the same day based on his current schedule.

This problem requires one to find the lowest common multiple of 2 numbers; in other words, \(lcm(6,12)=?\)

This one might be more obvious than other problems because 12 is a multiple of 6, and 12 is, of course, a multiple of itself. Therefore, \(lcm(6,12)=12\)

The result of \(lcm(6,12)\) is the fewest number of days until Matthew does both kinds of excerises again.

Here is a picture of a sample pyramid: 

Wait a second! \(\frac{14}{5}\) is an improper fraction already, so there is no need to change it.

\(2\frac{4}{5}\) is a mixed fraction.

My guess is that the guest misspoke.

\(22+222=244\)

Here's another method to solve this problem:

Not all authorities agree on the standard form of a line; some believe it to be \(Ax+By=C\) and others believe it to be \(y=mx+b\). However, for the purposes of this problem, I will utilize the first form, \(Ax+By=C\).

For a system, however, we have this form:
 

1. \(A_1x+B_1y=C_1\)

2. \(A_2x+B_2y=C_2\)

Let's convert both equations into that form:
 

1. \(y=\frac{1}{3}x-2\)

2. \(-2x+6y=18\)

Luckily for us, this equation is already in its desired form, so we can move on.

Ok, let's look at the equations side by side again:

1. \(\frac{1}{3}x-y=2\)

2. \(-2x+6y=18\)

Now, look at the ratio of the coefficients. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), then the system has no solution. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\), then the system has infinitely many solutions. If \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\), then there is exactly one solution.

Let's try to see if this works.

Let's check to see if \(\frac{\frac{1}{3}}{-2}=\frac{-1}{6}\neq\frac{2}{18}\)

First, let's simplify \(\frac{\frac{1}{3}}{-2}\)

Ok, let's check the condition again: Does \(\frac{-1}{6}=\frac{-1}{6}\neq\frac{2}{18}\)? Yes, 2/18 simplifies to 1/9, which is not equal to -1/6. 

As I stated above, when a system of equations has the relationship of \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\) then there is no solution. Therefore, this system has no solution. If there is no solution to this system, then the lines must be parallel because otherwise the lines would intersect and would have a solution.

Despite all of this work that I have shown, Hecticlar's method appears to be easier to follow and is probably more efficient because this method cannot tell if the lines are perpendicular...

To solve for x, you must use the square root and remember to split the answers into 2:
 

To solve for x and y, first solve for a variable and plug it into the other equation. Since you specified solving by substitution, I'll use that method. Here are your 2 equations:

1. \(2x+4y=-32\)

2. \(-3x+y=6\)

I'll solve for y in the second equation because it has a coefficient of 1.

Now that I have solved for y in one equation, substitute y in the other.

Plug x into an equation and solve for y. I'll plug it into equation 2

Therefore, the solution set is (-4,-6)

That's a determined child...

In this equation, our goal is to isolate the unknown, x. Let's see how we would go about doing that

There ya go!