Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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ganesh wrote:

Hi,

Of late, there has been a flooding of posts, only from newer members. I had to work 20 of 24 hours, and I had to curtail this. Many such irrelevant posts have tome to notice.

Often, I have to delete adult/irrelevant contents.

I don't have anything against. As an administrator (one of the three), Deleting messages is a last resort.

I am human too. There is every possibility I made a mistake. Working overtime.

Too many posts in a short time - This particular member does this often.

I wish to tell, normally, keep post count to 5 or less, as Administrators/Moderators have to scrutinize every post.

Ignore more threads moving forward. Stop deleting my questions. I want no interaction with you whatever. How dare you delete my questions after it takes me forever to put it all together using my cell phone!! You have been reported to Bob for the second time. Again, DO NOT answer my questions. Your answers will simply be ignored.

Bob wrote:

All hail the laptop. Better all round. But not so portable. Airport security is particular about anything above phone size. Schiphol; did I mention don't use it as you spend your life queuing; wanted me to remove the handkerchief from my pocket so they could scan it. WHAT!

I followed the link. We seem to have already had this conversation. Worryingly, I don't remember a thing from that thread. I'm hoping Imgur will come up with a bcc update soon.

Bob

Bob,

Someone is deleting my threads. I think it's ganesh who is upset that I reported him to you a few days back. If my questions are going to be deleted, then maybe I should leave the skte.

I posted two problems tonight:

1. An exponential equation problem.

2. A distance formula problem.

I cannot find my two most recent threads.

Who is deleting my questions?

phrontister wrote:

harpazo1965 wrote:Please copy and paste the following link. Tell me if you can see the image.

https://imgur.com/gallery/PXM8kTT

Yes, I can...as part of the following Imgur web page (a screen capture):

https://i.imgur.com/n3c7hgA.png

However, while your link works for me now, I've found from your other posts that your web page links only work for a limited time, if at all.

Here is the solution:

1. I will post problems that do not involve images.

2. I will post mostly word problems moving forward.

3. If I decide to post algebra, all work will be typed.

4. Let's move on beyond this topic.

5. Live long and love math!!

Bob wrote:

hi everyone,

My apologies for not being able to post properly the other day. I was away from home just using a Samsung A21s mobile and a hotel wifi. I don't like to openly state that I'm not at home so I had to be cryptic. After a two hour queue (no exageration) at Schiphol airport in the line for security (the start wasn't even in the building!) I am now back on my laptop.

I see there have been developments. The first post gave a link which didn't work on my mobile. But it did for Phrontister and he was able to convert so that it showed in his post. So we were part way to a method to get images . But that link now gives a 'no such address' error. Either harpazo1965 has deleted it or imgur have moved it. Odd.

Let's not waste any more time on that one. I answered the question in post 4 and the definition, if you really want one, would have to be -∞ < x < +∞.

That's like saying x is a real number such that x is a real number. True but not really adding anything useful.

I noticed while away that there are some more posts with images so I'm going to switch to one of those and see if I can get the image showing.

Bob

ps. How many posts are you allowed to make? I don't think there is a limit, but please use some common sense. If you post lots of questions I won't have time to answer them all. If you post too much chat, you're in danger of breaking the spam rule.

I will concentrate on word problems and questions that do not involve images moving forward. The image problem has been solved.

Bob wrote:

hi imcute

As you are new you perhaps won't have been following the difficulty here over a number of threads. On a previous server it was posssible to upload images directly to MIF but difficulties with that server meant the forum had to swiitch over and the direct upload facility was lost.

I use Imgur to get images here using a bcc code address from Imgur. I advised harpazo1965 to do the same. He signed up and uploaded his image there but the site does not support bcc on their mobile app, so this hasn't helped him.

If you know a way to get an Imgur image into bcc format on a mobile we'd all be really pleased to know it. Or maybe you know another way for him to get images here. Thanks.

Bob

Don't go crazy with this situation. I simply will use the forum for problems that do not involve images. For example, word problems will be my main concentration here.

No images needed to post word problems.

Bob wrote:

Tried this and I get an Imgur 404 error.

Bob

Sorry but this image thing is not easy to solve.

Bob wrote:

I am trying Phrontister's method for creating an Imgur image.

step 1. On my mobile, log in to Imgur and select an image.

step 2. Bottom right of screen, click the arrow and choose copy URL.

This is that url.

https://imgur.com/a/BURsC8w

If I enter it into my laptop browser address bar I can view the image. I'm not logged in to Imgur on my laptop so I assume anyone can do this.

step 3. On my laptop right click the image and choose 'copy image link'

step 4. I then pasted that back into this post. Here is the link:

https://i.imgur.com/7RjDiAX.jpeg

Note. It is different to the above and has jpeg as an extension.

step 5. Add bcc code to make the image appear. open square bracket, img close square bracket; image link address;open square bracket /img close bracket. Here is the resulting image.

https://i.imgur.com/7RjDiAX.jpeg

So, if a member is limited to a mobile I can convert their image so it appears in their post. They would have to do steps 1 and 2, and I'd have to do the rest.

Remaining question. Can steps 3 - 5 be done on a mobile? Still working on that.

Bob

This is only an image test. Please copy and paste the following link. Tell me if you can see the image.

https://imgur.com/gallery/PXM8kTT

phrontister wrote:

harpazo1965 wrote:Do you understand my question here?

Sorry, but I don't.

My last year of maths was 4th year high school and I don't recall ever covering these kinds of problems. So I guess they were taught in higher levels...or maybe I just dozed off in class.

1. The problems I have posted here thus far are intermediate algebra level. Nothing crazy like linear algebra or abstract algebra.

2. If you don't recall high school mathematics, why bother to reply? Seems like a big waste of your time and mine.

Abbey78336 wrote:

2x^2 + (432/49x) = 3

1. This is a game for you.

2. You are looking to see how many get it right and wrong.

3. What is this nonsense that 98% get it wrong? Are you among the 98% who get it wrong?

4. What have you done? Where's your effort?

Let me see.

2x^2 + (432/49x) = 3

2x^2 + (432/49x) - 3 = 0

The middle term can be expressed as (432/49)(x).

Use the quadratic formula. In the formula, a = 2, b = (432/49) and c = -3.

Take it from here.

imcute wrote:

mann

we could not answer your question or get your photo if you still use imgur urls

learn to upload

Thanks for letting me know. I will explore imgur a bit more on my next set of days off.

phrontister wrote:

Copy. Good one. Now, let's get back to my thread. Do you understand my question here?

Bob wrote:

I cannot explore any at the moment. I'll work on these in a couple of days.

Bob

No problem.

Question:

How many threads am I allow to post daily?

Bob wrote:

That should work but I still think the graph is the key. Once you have that the algebra is easy.

Bob

Graphing is ok but I am more interested in algebra manipulation.

phrontister wrote:

Thanks, Bob...glad it works for you.

But the OP said:

harpazo1965 wrote:Thanks but I don't have a computer or a laptop to do all that fancy stuff you talked about.

What device did you try it with, Bob? I'm hoping it was a smartphone, because, although I don't have one of those and can't perform the test, I thought my method would've been non-fancy enough to even work on a smartphone.

They have the capacity to perform the kind of copy/paste I referred to, don't they?

What are the steps needed for me to post photos here using my cell phone?

**harpazo1965**- Replies: 0

Copy and paste to see image.

https://imgur.com/a/PXM8kTT

**harpazo1965**- Replies: 0

Copy and paste to see image.

https://imgur.com/a/EGw3Xmd

**harpazo1965**- Replies: 4

Copy an paste to see image with question.

https://imgur.com/a/M32N6tL

**harpazo1965**- Replies: 2

Copy and paste to see image.

https://imgur.com/a/KVgLmrl

Bob wrote:

It's got a vertical asymptote at x=3. And a horizontal one at y=1.

If you attempt to solve for y=1 you get no x value.

So that's a clue. Try to construct the graph

Bob

The whole idea here is to find the range algebraically not using a graph. I don't recall much about asymptotes. Asymptotes is still a few sections away in my self-study of precalculus. Simply looking for an algebraic solution. Someone hinted that the domain of the inverse function is the range of the given function. Is that true?

1. Find the inverse of the given function.

2. Find the domain of the inverse function found.

You say?

navyboy484 wrote:

What’s your favorite subject?

Personally I’m really into math and scienceWhat about you?

Former Navy here (1996-2000). Are you in the Navy now?

ganesh wrote:

The equation

has two "obvious" solutions: x=2 and x=4. They are "obvious" because they are relatively easy to find by sheer guessing or by sketching the graphs of the two functions involved, and once guessed they are trivial to verify.https://useruploads.socratic.org/apnbo8bVTC6ioBCx7wzf_AutoGraph.jpg

There is a third solution with x<0. That this solution exists is fairly obvious as well: just consider the general behavior of the two functions for negative values of x and you will see that they must intersect at least once - in fact it's not hard to see that they intersect exactly once in that region.

However, it is not possible (we believe!) to write down an explicit formula for that solution by using rational numbers, arithmetic operations, and the so-called Elementary Functions: exponentials, logarithms and trigonometric functions. It is possible to write down such an explicit expression using a decidedly less standard function, the Lambert W function, and that's pretty straightforward too.

Proving the negative assertion above is far from trivial, and in fact I don't think that such a proof is known. Just because the solution can be expressed by evaluating W at ln(2)/2 (and making a few other simple calculations) doesn't rule out the possibility that there is also another expression for that same solution using elementary functions evaluated at rational numbers.

There's a brilliant and readable paper by Tim Chow on this topic, called "What is a closed form number?" . I highly recommend it to anyone interested in gaining a deeper understanding of those questions. It is a little surprising that many people wouldn't hesitate to make the assertion I made above, about the negative solution being impossible to write down "in closed form", without having proof of that fact and often without even being able to express precisely what the claim is.

We can see there is one solution in the interval -1 < x < 0 and a second solution at what appears to be x = 2 and a third at what appears to be x = 4. The two solution x = 2, 4 can easily be verified to be exact by substitution.

We can find the third solution numerically, using Newton-Rhapson method

Let

and using the Newton-Rhapson method we use the following iterative sequence.

And we conclude that the remaining solution is

to 4 dp.

A very detailed reply. I thank you also for the photo.

phrontister wrote:

Hi Bob;

I use a PC, and so maybe my experience differs from how the OP's device works (or can work) with Imgur, but this is what I've observed and got:

The OP's address isn't to the image itself...it's only to the Imgur web page with the image on it (amongst other things).

An image address has an image format extension, such as PNG, JPEG...which isn't the case with the OP's 'https://imgur.com/a/MMz2Kir'.

Solution:

I copied the OP's url into my browser's address bar, right-clicked the image that appears on the resulting Imgur page, selected 'Copy image address' from the drop-down menu, and then pasted that address into MIF (in 'img' tags nested inside 'hide' tags)...which gives me:

https://i.imgur.com/CRkvwKf.jpegYou'll see that the image address is very different from that of the OP's.

Maybe my method would work for the OP on his device...

Thanks but I don't have a computer or a laptop to do all that fancy stuff you talked about.

Bob wrote:

Many thanks, Phro. That's just what I needed. I'm gong to take out your hide tags to see if the image shows as was wanted. If so, you have found a way to get the image onto MIF from imgur. Yay! It works!

As to the poster's question its simply this. All x lies between minus infinity and plus infinity so why waste space including an inequality.

Bob

Thanks. I still would to see a defining inequality. Curious about what looks like.

**harpazo1965**- Replies: 4

Find the range of f(x) = (x + 2)/(x - 3) algebraically.

How we find the range of this rational function without a calculator?

**harpazo1965**- Replies: 2

A friend texted the following question to me.

Solve x^(2) = 2^(x) for x.

Any suggestions?