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Bob wrote:

hi mathland

2. I have decided to return to chapter 1 in my precalculus textbook. I want a complete review of the course. Are you willing to take the ride with me?

I am willing to help but subject to my conditions outlined below.

1. Did you ban me from the site two weeks ago?

Yes. You were warned in this post:

http://www.mathisfunforum.com/viewtopic … 53#p419353

On 15th May 2021 you posted 17 times. 8 of those posts were not related directly to maths. I note that no member has responded to any of them. This shows me that those posts were out of line with what the membership wants.

I am willing to help you to improve your mathematical ability but I was finding it impossible because you were flooding me with new requests faster than I could deal with them, and I was finding it very difficult to find your last post on a topic I was helping with.

After the warning you posted 4 more times in quick succession. I deleted 3 of the posts and kept one open. That one requires you to do some work and post some answers. That still hasn't happened.

4. Why would anyone ban me from this site? All I do is contribute to make the site better and more interesting.

That's your opinion. From the complaints I have received it is clear that others do not share it. The ban was temporary as it is my hope that you will post more reasonably in the future.

3. How many questions can I post per week to prevent getting banned again?

I'm a maths teacher with many years of experience. In class, when starting a new topic, I start by checking the pupils understanding of any preceding work, then launch into an explanation of the new topic, I intersperse this with many question and answer sessions. This is so I can check that what I'm saying is understood by the pupils. I may pick on a particular pupil when doing this if I think he is not paying attention. Then I set some exercises and check progress as I go around the class. Finally I set some homework so I can further check that each child has understood what to do. The results of this help me to decide whether to progress to the next stage or whether to spend more time on the topic.

Look at how many times I check progress. It's an essential part of the learning process. The pupils don't learn from me; they learn by doing it themselves (see Galileo quote in my signature)

I have tried to help you on a number of occasions. This post is the first time I've seen any of your work in response. It's a good start! Here's a reminder of what I said on 15th:

(1) Limit your posts to one per day.

(2) Make sure the post is about maths.

(3) Make sure a thread reaches a conclusion where you post a correct answer to your question and thank your helper(s).

Once a conclusion is reached then you are ready to post on a new question.

5. still have no clue how to upload pictures here. I need the steps ONE AT A TIME.

Step 1. Open an account at www.imgur.com

Say when you have done this and I'll give you step 2.

Bob

1. It was you who banned me for silly reasons. I did not see the warning you are talking about. This changes everything.

2.You are willing to help with precalculus subject to your rules. Interesting.

3. Make sure a thread reaches a conclusion. Are you saying threads are left unanswered?

4. My posts have not been offensive. My posts are not unreasonable. My posts are not threatening. Most of my threads in the DARK FORUM are about Christianity. What's wrong with the words of Jesus? In a world that is so cruel, so evil, I expect Bible verses and comments made by famous Christians like John F. Walvoord not to be offensive.

5. You are now saying that uploading pictures will be done one step at a time.

I don't know what to say. Searching for a math forum has not been easy. People dislike me everywhere I post questions.I have been banned from many forums not just this one. I have some serious thinking to do.

Bob wrote:

hi mathland

Original question: 18 yards from the endline and 12 yards from the

sideline. A teammate who is 42 yards from the same

endline and 50 yards from the same sideline receives ...You need to be clear about which number is an 'x' and which a 'y'.

I sketched a pitch with the pitch running left to right with one corner as the origin. Distances from the end line (y axis) are 18 and 42. Distances from the sideline (x axis) are 12 and 50.

So the correct coordinates are (18,12) and (42,50). And the correct pythag is:

So your original calculation was correct. But when you used the formula you had the x and y coordinates muddled so no wonder you got a different answer.

When I plug the points (12, 18) and (42, 50) into the distance formula,

Advice for next time: Always sketch a diagram and mark on the points. This will reduce your chances of an error.

Bob

Understood.

Questions:

1. Did you ban me from the site two weeks ago?

2. I have decided to return to chapter 1 in my precalculus textbook. I want a complete review of the course. Are you willing to take the ride with me?

3. How many questions can I post per week to prevent getting banned again?

4. Why would anyone ban me from this site? All I do is contribute to make the site better and more interesting.

5. I still have no clue how to upload pictures here. I need the steps ONE AT A TIME.

phrontister wrote:

The measurements given in the question are distances, not 'points' (exact locations) as such, and the error is in their placement within the 'points' notation (ie, the x,y elements in the brackets).

Care to clarify the error?

phrontister wrote:

mathland wrote:...points (12, 18) and (42, 50)...

It looks like there's an error in one of those points.

No error. The distances are as given in the question.

Bob wrote:

yes.

What about this set up?

d= √((42-18)^2+(50-12)^2)

d= √(24^2+38^2)

d= √(576+1444)

d= √2020

d is about 44.9 yds

When I plug the points (12, 18) and (42, 50) into the distance formula, I get a different distance for the pass. I get 2•sqrt{481}.

Which is correct?

Thanks

P. S. Did you ban me from this site about two weeks ago?

Bob wrote:

Look it up on Wikipedia.

B

Online sources use math jargon to explain this law. No problem. I will also look it up on You Tube.

Bob wrote:

hi 666 bro

I know you're trying to self-study so I'll give you the lesson I have used in the past for my A level students. It'll be longer than just answering your question, but I think you'll find it covers lots of topics that you are interested in. For my class it would take more than one lesson to do all this and, of course, if I said something a student didn't follow, they could ask questions and get me to go back over earlier topics. So take your time; do as much as you can in one session; and post back if you have any questions.

The lesson starts with "What does the graph of y = 2^x look like? I suggest you try this. Make up a table of values for y when x = -3, -2, -1, 0, 1, 2, 3 and plot the points.

Then repeat for y = 3^x and consider other values of graphs of the form y = a^x. I'll let you try this and leave a big space before adding my graphs.

https://i.imgur.com/W3ycKiy.gif

Graphs of the form y = a^x form a 'family' of similar curves. They all have a similar shape and all go through (0,1). As 'a' gets bigger the graphs go up more steeply

Next step. What is the gradient at (0,1)

Form an expression for the chord joining (0,1) to (Δx , a^Δx) and let Δx tend to 0.

Now, at this stage, we've got no way to say what 'k' is but later we'll find out. One thing should be clear; each value of 'a' will have a limit because we can see that the gradient at (0,1) exists, and is different for each 'a' value, and gets bigger as 'a' gets bigger.

Now the general gradient at any point on the curve.

So this family of curves has a very interesting property. The gradient function for each is the function again multiplied by the gradient at (0,1).

Next step.

So can we find an 'a' so that k = 1. This would mean that function differentiates to give itself!

Looking at the graphs we can see that different 'a' values give different 'k' values, so let's assume that such a value of 'a' exists. As it's special I'll call that one 'e' rather than 'a'. In other words

So e^x is a function that differentiates to give itself. But is it the only function with this property?

Let's say there are two such functions, f(x) and g(x). So df/dx = f and dg/dx = g.

Consider the function h = f/g and differentiate it. This needs the quotient rule.

If h differentiates to zero then h is a constant so f is a multiple of g. There are many functions that differentiate to give themselves back, but they're all multiples of each other.

Next step. So can we find a function that just has powers of x that differentiates to give itself back. (This is the answer to your question!)

I tell the class that the function has a '1' in it so it looks like this:

But what differentiates to give that 1? Answer x

So the function must look like this:

But what differentiates to give that x ? Answer x^2/2

So the function must look like this:

But what diffentiates to give x^2/2 ? Answer x^3 / 6

So the function must look like this:

Hopefully one of my class now says "That means it goes on for ever!". It's always great when they make their own discoveries like this.

So we carry on adding terms and also find an expression for the general term:

As many function exist that differentiate to give themselves this doesn't yet prove that e^x and that power series are the same but that'll come later. Looking at the a^x graphs it certainly comes at the right place in the family because you can work out the gradient for 2^x and it's under 1; and the gradient for 3^x and it's over 1, so our special 'e' number must be between 2 and 3. If you put x = 1 in the power series you'll get the value of 'e' and it lies in that range.

There is more but I'll take a break for now,

Bob

Continue with your notes. Interesting so far.

Bob wrote:

Please reply in that thread.

B

Ok.

Bob wrote:

hi mathland

Somehow I knew in advance that you would want to know about this so I searched back through all my posts and found this:

http://www.mathisfunforum.com/viewtopic.php?id=25550

My last post in the thread was an offer to finish the lessons but only if the guy was interested. I got no response then, but the offer still stands.

Reply on that thread please rather than this one.

Bob

Please, finish the lesson. I am interested. This is a lot to digest but not if I divide the reading into 2 or 3 days.

Bob wrote:

Are you talking about the sine rule and the cosine rule?

This link shows what the first is.

https://www.mathsisfun.com/algebra/trig-sine-law.html

If that's what you mean (I know you are familiar with it, just trying to make sure we're talking about the same thing) then I'll attempt to answer your original question.

Bob

Hi Bob. How are you? I am specifically talking about the law of tangents. I think there is a law of tangents. In fact, I know there is a law of tangents.

1. When do we use the law of tangents?

2. Why isn't this law discussed in trigonometry textbooks?

**mathland**- Replies: 1

I think MIF should have a Standardized Exams forum. The forum would be specifically for prep book questions. I am thinking about SAT, GRE, GMAT, ASVAB, etc. You say?

**mathland**- Replies: 0

I think MIF should have a forum specifically for math proofs. You say?

**mathland**- Replies: 0

Are you a NYC Teacher? If so, I would like for you to share the positive and negative aspects of teaching in NYC public schools before and during the pandemic.

**mathland**- Replies: 0

Anyone here works as a substitute teacher? I did for 8 years. Looking for people who are interested in discussing sub teacher stories, tips and experience.

666 bro wrote:

Now, I'm highly got interested in learning geometry by playing a game called "euclidea" but I forgot all the basics that I've learnt in my school so, where should I start learning?

Any resources and suggestions always welcome.

Are you still interested in learning geometry one chapter at a time?

**mathland**- Replies: 4

I know there is a law of sines and law of cosine in trigonometry.

What about the law of tangents? Why is the law of tangents ever used in trigonometry textbooks?

Hostible wrote:

Know that -a^2 means -1 x a^2 and not -a x -a.

The reason is that -a^2 has its coefficient as -1

Therefore -a x -a = a^2 and and not -a^2.

This symbol ^ implies exponent.

True but (-a)^2 is totally different.

We know that (-a)^2 = (-a)(-a) = a^2.

How about -(a)^2?

Let me see.

-[(a)(a)]

-(a^2)

-a^2

Agree?

**mathland**- Replies: 0

His disciples gathered in

the Upper Room that day,

to await the Spirit that was coming down;

Like I rushing, mighty wind,

it filled each one of them within.

I never knew how it felt, but I know now.

I never knew how it felt, but I know now.

I never knew how it felt, but I know now.

I never knew amazing grace

could find me right here in this place.

I never knew how it felt, but I know now.

Kneeling on my knees

I asked Him to please help me.

Take my cup, Lord fill it up, please do it now.

Then it started in my feet,

filled me up full and complete.

I never knew how it felt, but I know now.

I never knew how it felt, but I know now.

I never knew how it felt, but I know now.

I never knew amazing grace

could find me right here in this place.

I never knew how it felt, but I know now.

Oh, I never knew how it felt, but I know now.

Laverne Tripp

Blue Ridge Quartet

zetafunc wrote:

It might help to consider what we actually mean by e^x.

What is the definition of e^x?

1. The hint does not help.

2. Can you explain in MATH FOR DUMMIES style of language?

**mathland**- Replies: 0

Given . (x + y^2)^3 = 3x, find dy/dx. Then evaluate the slope m at x = 4.

Let me see.

I see the chain rule here.

Yes?

I know that dy/dx = m = slope.

Yes?

After finding dy/dx = m, I must evaluate the slope at x = 4.

Yes?

**mathland**- Replies: 0

Given ye^(x) = y − x, find dy/dx.

Questions:

1. dy/dx = y '

Yes?

2. dy/dx = y ' = m = slope

Yes?

3. For the given function, the product rule applies.

Yes?

**mathland**- Replies: 0

Find y ' and y "

y = rt{4 − x^2}

Let rt = square root

Let me see.

I must find the first and second derivative of the square root function.

Note:

I think the best way to tackle this problem is to

rewrite the given function as a power of x.

Doing this, I get y = (4 - x^2)^(1/2).

I now can apply the chain rule.

Yes?

Considering that the original function was given as a square root, must

I write the final answer as a square root? Is that mandatory?

Thanks

**mathland**- Replies: 0

Given e^(rt{x^2- 9}), find y prime.

Note: rt = square root

Let me see.

d/dx [e^(rt{x^2- 9})] = e^(rt{x^2- 9}) • d/dx [rt{x^2 - 9)]

I must differentiate the radical as the next step. Yes?

**mathland**- Replies: 6

I know that d/dx [e^(x)] = e^(x).

My question:

Why is this the case?

Why is e^(x) its own derivative?

Bob wrote:

Further notes on this:

cuberoot(t+4) is cuberoot(t) translated 4 units left in a horizontal direction.

Examine y = (x)^(1/3)

This graph and its inverse are 1:1 functions so we can proceed to raise both sides to the power 3

y^3 = x

We can reflect the graph of this function in the line y=x by swapping the x and y terms.

y = x^3

We know all about this cubic already. Both its domain and range are (-∞,∞)

The reflection interchanges domain and range. In other examples this might matter but as both are (-∞,∞) it doesn't matter.

The translation might effect some domains, but again, not this one.

So we can get domain and range just by considering simple transformations of the graph.

General hint. Cubics in general have certain common properties. We can call all such curves a 'family of curves' and use general properties to learn about new members.

Cubics with a positive x^3 term all start in the 3rd quadrant and finish in the 1st. If x^3 is negative this becomes 2nd and 4th quadrants.

Cubics have two stationary points (max and min) unless the two points merge into one in which case we have a point of inflexion.

Here are some general properties for quadratics:

All have a line of symmetry x = -b/2a . This line goes through the stationary point which is a minimum when the x^2 term is positive and a maximum when the x^2 term is negative.

Graphs of the form y = a^x

These all go through (0,1)

The gradient function is k . a^x

When k = 1, we use 'e' , making a function that has the interesting property that dy/dx = y.

I'm sure I did a thread on this but I cannot find it now using search.

Bob

Great notes. I will look further into this information and get back to you, if need be. More math questions posted later on today and tomorrow, Lord willing.