Membres: ilorty

If you really don't get it, let me tell you one thing. Since this is multiple choice, use trial and error. If you are on a test, nobody can help you with anything. So, if you don't get a multiple choice math question (or grammar) use trial and error.

But that would be pointless if I used it, because I know how to solve it, and you won't learn.

To begin, we distribute the 2 in 2(x + 7)

5x + 2(x + 7) = 14x – 7

5x + 2x + 14 = 14x -7

7x + 14 = 14x -7

Now, we need to isolate the x. First, subtract 14x from both sides. Then, subtract 14 from both sides.

7x + 14 = 14x -7

-7x + 14 = -7

-7x = -21

x = 3

3. 12t – 10 = 14t + 2. Find t

a. -6
b. -4
c. 4
d. 6

Here is what I don't get. If you could solve one, then you can solve 3. If you could solve 4, then you can solve 2.

Anyways, to begin, we want to isolate the t. The first natural process it to subtract 12t from both sides:

12t – 10 = 14t + 2

-10 = 2t + 2

Subtract 2 from both sides:

-12 = 2t

Divide by 2:

-6 = t

I solved both for you. Next time you post more than one question, know that you won't get a reply. (most likely)

I hope you understand now.

ilorty3 sept. 2020

+1084

Thank you Melody!

I just (okay, a few hours ago) renamed the thread into "what is happening 5". So if anyone is confused, just know that they are the same thing

ilorty3 sept. 2020

+1084

I used a google doc to type it, and it corrected a ton of my mistakes....but I can't seem to see the word "aloud" in my thread...let me check again

I am currently gathering materials for today's post. Your reminder helped me spot a mistake that google could not find, lol.

ilorty3 sept. 2020

+1084

Oops, my spelling is horrible!

Once, I was writing an English paper, and after seven rounds of editing, I still managed to spell "no" like "know".

ilorty3 sept. 2020

+1084

September 2, 2020:

Happy September, lovely people of the forum! I can't believe August is over! This is the first wrap post. Before going on, please read the introduction. (original post) It will set some guidelines, and tell you what the heck I am doing, and who I even am.

Before going on, I would like to recognize this answer created today:

https://web2.0calc.com/questions/coordinates_39#r3

It was by Omi67, a moderator that usually hangs out on the german forum, and sometimes on the english one. (well, most of the time...the german forum seems to be way quieter than this one)

Please go head over to the question, read the answer, and give them a like.

Here we go:

Bugs

There have been a few bugs lately. You can’t edit your post once you preview, you can’t give a like, and so on.

I found a way to give someone a like, even though the system does not allow that.

There are two ways to do it:

Enter that person’s profile.

Go into that person’s profile, find the answer that they created, and give the like! (don’t go into the answer, stay in the profile)

Go to the answer tab

Go into the answer tab, find the answer that they created, and give them a like! (don’t go into the answer, stay in the answer tab)

News

A lot of people have been wondering where CPhill is. Whether that is on the leaderboard, or for stalking purposes. (If the latter, please stop immediately, because I’m not letting you into my home sweet home)

He has been taking a break for the Summer, maybe to find Roman Numeral 0! (it has been a while since he aborted that mission. Maybe someone will convince him to go back to searching)

For people who want to know where he is on the leaderboard, press the “down arrow”. Then his name will appear at the top. (Sorry Melody...he’s still there!)

Problems

Recently, (mostly guests) people have been taking a lot of pictures of their problems. But, it is a pain to answer them….. BECAUSE THE PICTURE RESOLUTION IS OVERWHELMINGLY BIG!!!! (sorry, got a little excited)

So please, adjust the image size. You can use this site: https://resizeimage.net/ (i have not tried it yet)

If one of your dimensions is over 700, then let me tell yea, it’s way too big. The best size is probably a max of 500.

Unanswered Problem of the Day:

There have been a lot of unanswered problems, mostly by guests, that have not been reposted yet…

So I will feature a very “interesting” problem I found:

https://web2.0calc.com/questions/counting-question_16

It isn’t too hard, so I will give you guys the chance to try it.

Math comic of the day:

created by xkcd, you can check their work out

And….this concludes out wrap. (hope i did not disappoint you, melody :P )

ilorty

Comment here on the Latern Thread, but DO NOT comment on this thread

ilorty2 sept. 2020

+1084

I think it is the wrong link... it leads to this exact page!

ilorty2 sept. 2020

+1084

@ SHADOWwolf, don't you see that the link brings you back to the same page? What is the point of the link if it only brings you to the same page? Who are you trying to help? Are you just a troll that asks questions and clogs up the forum with "not so useful" links and answers like my hair clogs the bathroom drain? Or is it the wrong link? (The preview is there for a reason)

Anyways, for your question we will need variables:

A = Amy's coins

B = Ben's coins

C= Carl's coins

D= Debbie's coins

With what we know, we can say this:

B = 3A

3B = C

3D = 2C

ABCD = 162

First and foremost, we need to write each variable in terms of one variable. Let's use A:

A = A

B = 3A

C = 3B = 3(3A) = 9A

D = 2/3C = 6A

Therefore:

ABCD = A (3A) (9A) (6A) = 162A^4 = 162

A = 1

Now that we know that A is 1, we can say that:

A = 1

B = 3(1) = 3

C = 9(1) = 9

D = 6(1) = 6

To check the work, let's multiply:

1 * 3 * 9 * 6 = 162.

so our answer is right.

ilorty2 sept. 2020

Nom d'utilisateur	ilorty
But	1084
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