Links: Terminating decimal and Recurring decimal.

Thanks for the links. I will check it out.

]]>https://imgur.com/gallery/6mjyDcd

]]>https://imgur.com/gallery/e6ySIqm

]]>The question looks easy enough but it's tricky to me.

I know there are 24 hours in a day. So, I decided to divide 12,996 by 24 hours, which yield the decimal number 541.54166666667.

Of course, this decimal number tells me nothing.

A. How do I find the correct answer?

B. If the quotient found leads to the right answer, can you show me how?

]]>Help me! I newbie!

Help you with what? We need a specific question with clear instructions. So, post your question(s) and wait for someone to reply.

]]>https://imgur.com/a/ZSarUyQ

]]>https://imgur.com/a/qtBeMMK

]]>amnkb wrote:"odd" means it matches on both sides of diagonal y = x

**

That's not the definition I have always used.

https://www.mathsisfun.com/algebra/func … -even.html

The url in post 3 worked when I made my post but not now. Maybe the poster removed it from imgur ???

There were two graphs. The first was y = x^2 or something similar. The second was, I think, y = e^x or some thing similar. There were no scales so the exact function could not be determined but enough to show the symmetry or not.

Bob

ps ** If a graph of a function is reflected in the line y=x this is equivalent to interchanging y and x and the resulting graph shows the inverse function.

eg. Compare y = 2x + 3 and y = (x-3)/2

Hi Bob.

Yes, I removed the photo from imgur. I will post several questions later showing my work or at least effort. I need all the practice I can get with imgur.

I am not savvy in terms of computers, laptops, tablets, and cell phones. I just know internet basics. Again, I deleted all imgur uploads yesterday. I will keep my uploaded stuff there moving forward.

harpazo1965 wrote:Bob wrote:The two expressions have p^2 in it and 2ap if you put a = 7

So add a^2 = 49 and you have a perfect square.Let me see.

(14/2)^2 = (7)^2 = 49

The missing number that completes the square is 49.

You say?yes

thatiswhat he said

Ok. Cool.

]]>harpazo1965 wrote:I am thinking of a number. It lies between 1 and 10; it's square is rational and also lies between 1 and 10. The number is larger than pi. Correct to 2 decimal places (that is, truncated to two decimal places) name the number.

ne # between these works

infinitely many others do 2

Thank you. Interesting question.

]]>harpazo1965 wrote:Suppose that m and n are positive integers with m > n.

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

e_jane_aran wrote:I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions.

Which pairworks?harpazo1965 wrote:

If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.If I do that, I am left with m^4 + n^4 twice. In other words, I get

2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)

If I add a^2 + b^2 I get c^2. So, the addition of the two legs of this triangle

yields the value of the hypotenuse. You say?

Hello, I've purchased Brilliant app, and I've been wondering if anyone could answer me if after the month that I've paid for Premium, if I cancel my subscription for the next month, does the progress I've done on some courses stay? Like are those courses I've done while I had the premium subscription stay available to redo after subscription if I want to?

dunno if ne1 else here has used the app

Google Play: https://play.google.com/store/apps/deta … n_US&gl=US

Apple App Store: https://apps.apple.com/us/app/brilliant … d913335252

maybe u can contact brilliant & ask?

https://help.brilliant.org/en/collectio … ct-support

gud luck!

Thank you very much for digging that out. A picture is indeed worth a thousand words. ]]>