If you have a specific question about a topic that you would like to ask or have explained, post it as a comment to this thread. Then check in later for a response.

I'll check and update the comments periodically throughout the evening.

**Update #1: **

Anonymous said...
*
Could we go over completing the square for circles? Also, the equation of a circle?
*

Okay, "Anonymous". It's like this. The standard form of the equation of a circle is

**(x - h)**

^{2}+ (y - k)^{2}= r^{2}*r*is the radius.

Note: you can use the

**Distance formula**or Pythagorean Theorem to see if another point they give is on the circle if the distance from the center to the point is the same as the radius.

As you might remember from *Algebra*, (x - h)^{2} is the square of a binomial, and can be written as *(x - h)(x - h)*. If you were to multiply that (think FOIL), you would get *x ^{2} - 2hx + h^{2}*. That expression would be a

*perfect square*. If you have an incomplete expression, you need to complete it, so that you can factor it.

Say they give you *x ^{2} + y^{2} + 6y = 16*

x

^{2}is a perfect square. It's the same as (x - 0)

^{2}.

However, y

^{2}+ 6y is not a perfect square -- it doesn't have a constant term.

Do you see the

*2hx*in the expression above? The middle term of the complete square is

*double*the number in the binomial, so you need to find

*half*of it.

Half of 6 is 3. The final term of the completed square is

*h*, h is 3, so h

^{2}^{2}is 9.

So you need to add 9 to both sides of the equation, which gives you

*x*

^{2}+ y^{2}+ 6y + 9 = 16 + 9*x*

^{2}+ (y + 3)^{2}= 25In this case, the center is (0, -3) -- because you flip the sign -- and the radius is the square root of 25, which is 5.

It is quite possible -- likely, even -- that they will give you a problem with a radius that has an irrational length.

Does this help/answer your question?

**Update #2: **

Anonymous said...*
Can you go over density problems*

Density is mass divided by Volume. Imagine you have something that weighs 10 pounds. If it fits in the palm of your hand, it's pretty dense. If it's the size of your kitchen table, it's not very dense at all.

D = m / V, like those "dense" people you meet at the D.M.V. when you apply for a learner's permit.

They have to give you 2 of the three values, so that you can find the third one. However, they can make you figure out Volume.

Volume of a prism is the Area of its Base times its height.

A rectangle prism would be length X width X height

A cylinder would be pi * r^{2} * h, etc.

After that, it's likely to just be an Algebra problem.

I don't have a specific example I can give you or that.

**Update #3: **

Anonymous said...*
Can you go over proofs*

Not really. I could spend a week on proofs. If you want something specific, I would check my old Regents exam posts.

Here are some general guidelines.

Look at the image. What do you see? What do things look like? You **CANNOT** go based on looks, but it might give you a direction to go in.

Don't "assume" anything. Either it's given, or you derived it from what was given.

Make a plan. How are you going to get there?

Does it involve proving triangles are congruent? Then you'll need SSS, SAS, ASA, AAS or HL. (Don't make a backward SSA of yourself!)

If you use any of those, you need to specify three pairs of things that are congruent. In the case of HL, make sure you state that all right angles are congruent. Seriously -- it needs to be stated.

If you are proving that two sides of a triangle or two angles are congruent, then the last reason will probably be CPCTC (Corresponding parts of congruent triangles are congruent).

They won't give you anything that you can't figure out. Two of the biggies are vertical angles are congruent, and the reflexive property (for sides or angles).

If it's a circle, remember that you can add extra radii, and that all radii of a given circle are congruent. Tangents are perpendicular to the radius, so you might see a right triangle in the circle. Similar triangles (use AA) inside the circle are also possible.

Make sure you state all the given, and if there's an illustration, mark off everything you know. It might give you ideas, or it might remind you what you haven't explicitly stated yet.

Obviously, you need to know your theorems. And there are a lot of them.

Does this help/answer your question?

**Update #3: **

Heaven said...*
Could we go over finding a section of a circle? And those circle problems dealing with an external point?*

I assume you mean a "sector" of a circle, like a slice of pie? Think of slice of pi, if that helps.

The area of the sector of a circle is a fraction of the area of the entire circle.

The fraction that you need to multiply by is the central angle over 360 degrees. (Times pi r square)

They can also ask the reverse. They can tell you the area of the sector and the radius and have you come up with the central angle by working backward.

Think Algebra: inverse operations.

I'm not sure what you mean by "those circle problems dealing with an external point".

Do you mean finding the size of an angle from the arcs the lines intersect?

Do you mean the relationship between the lengths of the secants or tangents from an external point?

Do you have an example?

**Final Update ... It's Friday morning**

Anonymous Anonymous said...
*Can you go over finding a point on a circle, and also ratios of line segments?*

Suppose you are given an equation like (x - 3)^{2} + (y + 1)^{2} = 20.

If you wanted to know if a point is on the circle, say (5, 3), substitute those values into the equation.

(5 - 3)^{2} + (3 + 1)^{2} =?= 20

2^{2} + 4^{2} = 20 ?

4 + 16 = 20

20 = 20, check

Therefore, (5, 3) is a point on the circle. Had it equaled anything other than 20, it would not have been on the circle.

If a point in on a line somewhere that isn't the midpoint, you need to use ratios to find its position.

For example, if given A(-1, 2) and B(7, 8) and you and P was a point such that the ratio of the lengths AP:PB was 3:1, where would P be?

First, 3 + 1 = 4, so P is 3/4 of the way from A to B.

Find the difference of the x values 7 - (-1) = 8, multiply it by 3/4, and you get +6.

Find the difference of the y values 8 - 2 = 6, multiply it by 3/4, and you get +4.5

Add those values to the coordinates of A to get P. P(-1+6, 2+4.5) gives you P(5, 6.5).

Yes, you can get a decimal.

## 6 comments:

Could we go over completing the square for circles? Also, the equation of a circle?

Can you go over density problems

Can you go over proofs

Could we go over volume and density problems

Could we go over finding a section of a circle? And those circle problems dealing with an external point?

Can you go over finding a point on a circle, and also ratios of line segments?

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