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#1 Exercises » Circle Problems » Today 07:57:02

mathxyz
Replies: 0

1. The center of a circle is the point (3, 2). If the point ( 2, 10) lies on this circle, find the standard equation forthe circle. 54. Find the standard equation of the circle tangent to the x-axis and with center (3, 5).

Hint: First draw a sketch.


2. Find the standard equation of the circle tangent to the y-axis and with center (3, 5).

3. Find the standard equation of the circle passing through the origin and with center (3, 5).

4. The points A( 1, 6) and B(3, 2) are the endpoints of a diameter of a circle, as indicated in the accompanying figure. Find the y-intercepts of the circle.

Hint:Could you do the problem if you had the equation of the circle?

Enjoy!

#2 Re: This is Cool » A Point Cannot Be Measured » Today 07:52:52

KerimF wrote:

For instance, this is indeed one of the proofs that being non-existent for not being well-defined doesn't imply being non-useful in one's life.

A point cannot be measured is found in most if not all geometry textbooks.

#3 Re: This is Cool » A Point Cannot Be Measured » Today 07:51:26

Bob wrote:

The idea comes from Euclid.  He wrote out a set of geometry axioms for an idealised version of geometry.  He then showed how to derive many theorems  from these and wrote a book "The Elements" that has been used by mathematicians ever since.

Not only does it give us many famous results (eg the angles of a triangle add up to 180) but his axiomatic approach has given mathematicians the way to develop any mathematical theory.

He defined a point as having a position but no size.  Lines have length but no width.  It's an idealised view. When you mark a point on paper and look at it under magnification you can see your point does have size.

Bob

Thank you, Bob. Always very informative and educational replies.

#4 Jokes » George Washington Joke » Today 07:48:29

mathxyz
Replies: 0

Why did George Washington cross the Delaware River?

Answer: He wanted to get to the other side.

#5 Formulas » Math Formulas? Really? » Today 07:45:30

mathxyz
Replies: 0

Can I post math formulas here without my posts deleted?

#6 Maths Teaching Resources » Professor Leonard Video Lessons » Today 07:43:29

mathxyz
Replies: 0

I highly recommend watching professor Leonard on YouTube. This man was born to teach mathematics. He comes down to the level of students by reducing the complexity of abstract topics.

#7 Maths Is Fun - Suggestions and Comments » Suggestions? For what? » Today 07:39:10

mathxyz
Replies: 0

Why should I make suggestions to improve the site only to find my post deleted the next day by a ghost writer?

#8 Re: Science HQ » Is our planet dying? » Today 07:36:15

paulb203 wrote:

Is the planet (Earth) dying?
Was it ever alive?
Before the first soil, before the first moss, or moss-like plant, or blue-green algae - basically before the first life forms that would attach themselves to the surface of the Earth, what was the surface like?
Was it smooth, like an enormous (almost) spherical stone? What was it’s composition, approximately?
And, back to the top, if every life form on Earth went extinct and all the corpses (including the plant ‘corpses’) decomposed, to dust (?) would the planet be just as it was 4-5 billion years ago before life arose on it?

The planet Earth is dying. The condition is sin.

#9 Introductions » From Sunset Park Brooklyn NY » Today 07:35:05

mathxyz
Replies: 0

I will not disclose my name here. I live in Sunset Park Brooklyn NY.

A little bit about me:

1. I am 59 years old. Yeah, middle-aged and getting older quickly.

2. I have several degrees in areas other than mathematics.

3. I love math. I should have majored in mathematics.

4. I am a "somewhat" introvert. Love my alone time.

5. I work full-time as a security guard. A security guard with 3 college degrees. Now, you can laugh.

6. I live in NYC but hate what the city has become.

7. I play solo guitar at the intermediate level.

8. I read and compose music for solo guitar when time allows.

9. I joined this site to do a self-study of math learned in the 1980s and 1990s. Wow! I am getting old....

10. I am divorced and single. In fact, I have been single for over 2 decades.

#10 Re: Dark Discussions at Cafe Infinity » Message To Bob » Today 07:22:45

Bob wrote:

If you're crazy then I must be too.  Stick to what you want to do.

Bob

I am totally surprised that this post was not deleted by the hater(s).
I am confused. This forum is basically to share whatever we want that is or is not related to math. I don't get it.

#11 Re: Dark Discussions at Cafe Infinity » Bob...My Recent Posts » Today 07:20:42

Bob wrote:

Not me.  I did wonder about them though.  I count 6 missing. This forum has members from all over the world. That means a spread of politics and religion.  All posters need to think about this and take care with potentially controversial subjects.  But we do allow a wide range of topics so I was inclined to wait and see.  We had a poster a while back who started with an innocent statement but it provoked unpleasant responses so in the end we had to terminate it.

I recommend you stick to maths.

Bob

I got a pretty good idea who deleted my posts. It's the member who never replies to my posts.

#12 Re: Dark Discussions at Cafe Infinity » Back In My Substitute Teaching Days » Today 07:19:29

KerimF wrote:

I say... this may happen to anyone working under a direction. I learned it in the hard way.
I mean, my father died (when I was 9) also because of jealousy.
He was working in a French bank (a branch in Middle East). Being very good in math and French, he became, at age 37, the general manager of the other branches in Middle East too. Two years later, the oldest one among his personnel ended up, on one day (in 1958), to present false evidence (sent to France via telex) that my father is a gambler. The next morning, his bank received a telex from France that he was fired (this was very fast, everything was prepared in advance). My Father survived a few months later before his two kidneys stopped working for good.
This is why it became out of question for me to serve any rich families even for all treasures and pleasures of the world. Therefore, at school, then at the universities, I was looking always for good knowledge only, nothing but knowledge (yes, I was never interested in degrees), so that I can gain my daily bread (and of my assistants) while I am really free and independent (running my small private business related to electronics, since 1975 till now).


Thank you for sharing your story.

#13 Re: Help Me ! » Factor Completely » Today 07:17:47

Bob wrote:

(x - 1)(x + 1)(x^2 + 1)

Correct!

(x - 1)^2(x^2 + 1)

This is not the same. It is (x-1)(x-1))x^2+1)

Bob

Thanks.  So, factor completely means showing every factor.

#14 Re: Help Me ! » Factor Completely » Today 07:16:35

Bob wrote:

Nothing else can be done here.

3 linear factors! I'd say that is fully done.

Bob

The x^3 term tells me that there are 3 linear factors that must be found.

#15 Re: Help Me ! » Factor Completely » Today 07:14:40

Bob wrote:

Correct.

Bob

Just another review problem.

#16 Re: Help Me ! » Find Number & Factor » Today 07:13:10

Bob wrote:

That's it done nicely.

Bob

Thanks. This is one is slightly tricky because of the fraction.

#17 Re: Help Me ! » Find Number & Factor » Today 07:10:31

Bob wrote:

Yes, what you have done is correct.

Note: The question did not say to solve for p.

 

That's because the original was p^2 + 14p but it wasn't equal to anything so there wasn't actually an equation to solve.

Bob

Sullivan did not say to solve for the variable in this exercise. Like you said, the problem is not an equation

#18 Re: Help Me ! » Factor By Grouping » Today 07:08:30

Bob wrote:

Correct!

Bob

Don't get me wrong. I know how to factor. We are in the review section of the book.

#19 Re: Help Me ! » Factor Difference of Cubes » Today 07:06:56

Bob wrote:

Let 27 be 3^3. This means a = 3 in the formula.

Let 8x^3 be (2x)^3.

Good

But then

(2x - 3)[(2x)^2 - 3(2x) + 3^2]

Bob

Oh yes. I forgot the final step.

#20 Re: Help Me ! » Long Division » Today 07:05:53

Bob wrote:

Ok. First notice that putting x=a causes x^5 - a^5 to become 0, so (x-a) is a factor => no remainder.
       
                                       x^4       + a   x^3       +    a^2 x^2   + a^3 x      + a^4
                  _____________________________________________________
x - a           |   x^5     + 0 x^4       + 0   x^3       +     0    x^2       + 0   x     - a^5
                      x^5      - a x^4
                     ______________
                                    a x^4        + 0   x^3
                                    a x^4        -a^2 x^3
                                   ___________________
                                                   + a^2 x^3       +    0    x^2
                                                   + a^2 x^3       -    a^3 x^2
                                                   _________________________
                                                                          +    a^3 x^2      + 0    x
                                                                          +    a^3 x^2      - a^4 x
                                                                          _____________________
                                                                                                   + a^4 x    - a^5

                                                                                                   + a^4 x    -  a^5
                                                                                                   _______________

                                                              remainder              =         0              0

so x^5 - a^5 = (  x^4       + a   x^3       +    a^2 x^2   + a^3 x      + a^4)(x-a)

Bob


I must say that you really know how to do long division by trying on different spaces in the typing area.

#21 Re: Help Me ! » Long Division » Today 07:03:48

Bob wrote:

Easier .... like using the quadratic formula rather than completing the square.  But where did the formula come from?  Mathematicians like to know the answer to that question ... understanding rather than rote learning.

Bob


I totally get it. Sullivan wants students to practice long division on that particular page.

#22 Help Me ! » Factor Completely » Today 06:01:02

mathxyz
Replies: 2

Factor the polynomial completely.  If the polynomial cannot be factored, say it is prime.

Problem 98; page 58.

x^4 - 1

(x^2 - 1)(x^2 + 1)

The right side factor is prime which means that it cannot be further factored.

The left side factor becomes (x - 1)(x + 1).


Answer:

(x - 1)(x + 1)(x^2 + 1)


Question:


Can I also write the answer this way?


(x - 1)^2(x^2 + 1)


Enjoy your Sunday.

#23 Help Me ! » Factor Completely » Today 05:53:04

mathxyz
Replies: 2

Factor the polynomial completely.  If the polynomial cannot be factored, say it is prime.


Problem 90; page 58.


x^3 + 8x^2 - 20x


GCF = x


x(x^2 + 8x - 20)


x(x - 2)(x + 10)


Nothing else can be done here.


You say?

#24 Help Me ! » Factor Completely » Today 05:47:10

mathxyz
Replies: 2

On page 58 there is a big box that Sullivan calls Mixed Practice. I will spend several days inside this box reviewing factor completely. On that same page,  there is another series of factoring problems under the name Applications and Extensions.

For Applications and Extensions, there is a TRIANGLE FIGURE with an integral symbol in the middle. By this, Sullivan means that the next set of questions are often found in calculus. So, in short, I will be on page 58 for several days before moving on to Section R.6 or Synthetic Division. I will only select problems that I haven't factored completely in many years or that I simply like.


Page 58;  problems 84.


Factor the polynomial completely.  If the polynomial cannot be factored, say it is prime.


3x^2 - 12x + 15


Factor out GCF.


Our GCF = 3


3(x^2 - 4x + 5)


You say?

#25 Help Me ! » Find Number & Factor » Today 05:30:02

mathxyz
Replies: 2

Determine the number that should be added to complete the square of the expression. Then factor the expression.


Problem 74; page 57.


x^2 + (1/3)x


[(1/3) ÷ 2]^2 = (1/6)^2 = 1/36.


I now have x^2 + (1/6)x + (1/36) = (1/36).


The number I want is (1/36).


Now I factor x^2 + (1/6)x + 1/36.


(x + 1/6)(x + 1/6) = 1/36


(x + 1/6)^2


Answer:


The number that should be added to complete the square is 1/36.

The factored form is (x + 1/6)^2.


You say?

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