All Answers 358090 Answers

20, 20, 10
20, 20, 5, 5
20, 10, 10, 10
20, 10, 10, 5, 5
20, 10, 5, 5, 5, 5
20, 5, 5, 5, 5, 5, 5, 

10, 10, 10, 10, 10
10, 10, 10, 10, 5, 5
10, 10, 10, 5, 5, 5, 5
10, 10, 5, 5, 5, 5, 5, 5
10, 5, 5, 5, 5, 5, 5, 5, 5
5, 5, 5, 5, 5, 5, 5, 5, 5, 5
(So I believe 12 ways.)

That means "line segment 'CE' ".

http://duellplumbingandheating.com/10-common-toliet-problems-the-solutions/ 

Would this help?

 z² + 2z - 15 = 0

You can use the quadratic formula, complete the square, or just factor it by looking at it and telling what the factors are. The last one takes the most practice, but it is the fastest once you get the hang of it.

On this one I would just factor it, but since I don't know how to explain my thought process very well, I will just show you how to use the quadratic formula on it.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x = z

a, b, and c are the constants of each term.

a = 1, b = 2, c = -15

Just plug these into the formula.

\(z = {-2 \pm \sqrt{2^2-4(1)(-15)} \over 2(1)} \\~\\ z = {-2 \pm \sqrt{64} \over 2} \\~\\ z = {-2 \pm 8 \over 2} \\~\\ z = -1 \pm 4\)

z = -5  or  z = 3

625.318471338

Im not saying its going to get rid of the spam, its going to get rid of people that ask question that arent related to math but arent spammers.

And yes, they could enter reddit/tumblr/facebook/twitter/4chan etc etc but for some reason they dont. Only god knows why.

Basically, this website answers mostly math questions and provides a scientific calculator. Just click "Register" to register an account, and follow the given steps from there.

1.41421356237

Yes there is some spam on here, but it could be much worse. Then again, maybe the mods get to delete it before I have a chance to see it.

But would having a second forum really get rid of the spam that we do have? There are already plenty of forums on the internet that are made just for talking about anything, like reddit. If the spammers don't go to other already existing forums to post their question, what would make them go to another forum on this website to post their question?

Also, just to throw this out there, I really like how guests can post on this forum. I get annoyed with almost every single website asking me to "create an account" before I do anything. When I finally discovered this website that allowed guests to post, I was so surprised and happy. I wouldn't be on this forum if I wasn't able to make a post as a guest. I would have seen "make an account to post" and immediately turned away. However I understand that if the forum gets a lot bigger that would probably have to change.

*applause*

"!" Means that you time the number with the number that came before until you reach 1
With 4 it is: 4*3*2*1=24
 

What shape are you trying to find the height of?

What information do you have about that shape? (e.g. the area, volume, perimeter, surface area, or measure of an angle)

no          

-8n  = 38     divide both sides by -8

n  = -38 / 8  =  -19 / 4

It doesnt work. the only way i can see its not related to the topic is if an admin checks it and marks it. By that time they call spam more, and checking if a post is related or not can take time. you know that better than me. They wont spam if every spam post will result in deleting the poster's user.

and adding another forum is no big deal.

CONS:

none (The way i see it)

PROS:

saves time.

i think one of the worst thing a site can be is a human- dependent site.

And this is not the only way to do this, obviously, but i think its a way you can (and should) use.

Test it out if you're unsure. I have to do this a lot to check myself. Just pick some random numbers for a, b, and c and test it.

I pick a = 2, b = 3, and c = 4

2 ^ ( 3 * 4 ) = ( 2 ^ 3 ) ^ 4

2 ^ ( 12 ) = 8 ^ 4

4096 = 4096

So yes they are equal.

You can see why too:

(2³)⁴ = (2³)(2³)(2³)(2³) = (2)(2)(2) (2)(2)(2)​ (2)(2)(2)​ (2)(2)(2)​ = 2¹²

Or you can do this:

We know that 1/4=25% or 0.25, so 1/8 is 25%/2 which equals 12.5% or 0.125. Multiply this by 3 to get 0.375 or 37.5%. Both ways get accurate answers.

yeah, but what im saying is we can have a forum for non math subjects, so people who dont spam on purpose can post there.

And the only way to stop the spam is to delete the spammers.

i dont think math jokes or daily quotes are spam, and im not sure about people posting "good morning" every d**n morning or congratulating every new user, but we there are posts that are definitely spam.

this is not a forum for random things, its a math forum. If you want to talk about subjects that arent related to math you can simply enter the "no math forum" or however it should be called and post whatever you want.

and although i usually post as a guest, i think not allowing to post as a guest is an option.

Hi Ehrlich

you can hide all "off-topic" questions:

33 / 40   =  82.5%.........on a 10 point grading scale.....this is a "B"

3-x3x+2=? or 3-x*3*x+2=? or just -3x^2+5=?

is like saying A=B (aka: a Line) there are two variables, so

there's a infinite amount of solutions and all of those solutions make a line.

Go to google and type in the search bar:

     Graph -3x^2+5

And it should show you the line.

Good suggestions, Ehrlich, and we try to "white out"  the garbage.....unfortunately, there is no "spam" or "garbage" filter so whatever someone wants to post gets on.......this is probably true of most forums, unless you have "gateway" moderation

Some of the posts are innocuous [ MysticalJaycat's, for instance ] and we do encourage  some fun on here.......

My recommendation.???.....just ignore the gunk......we still have some pretty good questions being asked..... 

Hmmm, in situation 2, is x=3 really a solution?

I don't think it is, but I don't know what the " error " is in your work...

Is there a way to know that without just testing it in the end?

3

(a)   I do not know the proof, but there is a trig identity that says that 

a sin x   +  b cos x  = c sin (x + α)      where c =  sqrt (a^2 + b^2)   and

α = atan2(b,a)  

So

c =  sqrt (4^2 + 3^2)  = sqrt (25)   = 5    and

atan2 (b , a)   =  atan2( 3 , 4)  ≈  .6435     (in rads)

4 sin x  + 3 cos x   =   5 sin ( x + .6435 )

(b)  4sin x + 3 cos x  = 2  

4sin x    =   2 - 3 cos x        square both sides

16sin^2 x   =  4 - 12cosx  +   9cos^2 x

16(1 - cos^2 x)   =  4 -  12cos x   + 9cos^2 x

16 - 16 cos^2 x   = 4  - 12cos x  + 9cos^2 x

25 cos^2x - 12 cos x - 12  = 0        let cos x  = a

25a^2    -  12 a     -   12   = 0

Using the quad formula.......

a  = [ 6  ±  4 sqrt (21)] / 25

So  

a  = cos x  = [ 6  ±  4 sqrt (21)] / 25

Wnen    cos x  = [ 6  + 4 sqrt (21)] / 25

arccos [ [ 6  + 4 sqrt (21)] / 25 ]  = x  ≈   .232 rads   and  ≈ 6.051 rads

When  cos x  = arccos [ [ 6  - 4 sqrt (21)] / 25 ]

arccos [ [ 6  - 4 sqrt (21)] / 25 ]   = x ≈ 2.0866 rads    and  ≈ 5.228 rads

However.......232 rads  and 5.228 rads are extraneous  [ because of squaring ]

As the graph here shows.....the two solutions are   x ≈ 2.0866 rads  and  x ≈ 6.051  rads

https://www.desmos.com/calculator/pjvewx15zw