In order for a point to be on the line, the point must satisfy the equation given. Thus, plug in the  and  coordinates to see if they will give you a true equation.

If you plug in  into the equation, you will get the following:

Thus,  satisfies the equation and must be on the line.

Which of the following points is on the following line? 

So, to test this, we can plug in each choice and solve to see if they make sense. 

To save time, let's test the easier ones first. Recall that anything times 0 is 0, so we should try out the options with 0's first.

Recall that ordered pairs represent an x and a y value with the x coming first: (x,y)

So, this is not our answer. However, it does give us a hint as to the correct answer.

When we plugged in 0 for x, we got 6 on the right hand side. This means that if we plug in 0 for x, then we should get 6 for y. So, let's try out our next point.

So, our answer must be (0,6)

In order for a point to be on a line, the point must satisfy the equation. Plug in the values of  and  from the answer choices to see which one satisfies the equation.

Plugging in  will give the following:

Since  satisfies the given equation, it must be on the line.

Find the y-coordinate which would would make the slope between the following points equal to 5.

To find the slope of a line, use the following formula:

Now, we know all of these values except one. Let's find it.

So, our answer is

Therefore:

This is a quadratic equation in standard form, so first we need to factor .

This can be factored out as 

where .

By trial and error we find that , so

can be expressed as

.

Set each linear binomial equal to 0 and solve separately:

The solution set is .

The area of a right triangle with legs of length  and  is 

.

Substitute  and  for  and  and 600 for , then solve for :

We can now factor the quadratic expression:

Set each linear binomial to 0 and solve to get possible solutions:

Since  must be positive, we throw out the negative solution. 

.

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract  from both sides to get 

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for . To do this we must make a factor tree of  which is 28 in this case to find the possible solutions. The possible numbers are , , .

Since  is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal . So ,  , and 

Success! 14 plus 2 equals . We then plug our numbers into the factored form of 

We know that anything multiplied by 0 is equal to 0 so we plug in the numbers for  which make each equation equal to 0 so in this case .

The center is located at (3,4) which means the standard equation of a circle which is:

becomes

which equals to

Change division into multiplication by the reciprocal which gives us the following

Now 

this results in the following:

Simplification gives us

  which equals