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Ryan had some blue and green buttons. 3/7 of the buttons are blue and the rest were green. He gave away 2/5 of his buttons. 80 of the buttons given away were blue. 2/3 of the buttons left were green. How many buttons did Ryan have altogether at first?

(3/7)  are blue    (4/7)  are green....let N  be  the total

No.remaining  = blue remaining + green remaining

(3/5)N =   [(3/7)N -80 ]  +  (2/3) (N - (2/5)N)        simplify

(3/5)N  = (3/7)N  - 80  + (2/3)N  - (4/15)N

(3/5)N  =  (3/7)N  - 80  + (10/15)N - (4/15)N

(3/5)N  = (3/7)N - 80  +    (2/5)N   

80 = (3/7)N   + (2/5)N - (3/5)N                                         

80 =  (3/7)N  - (1/5)N                                                                7                                                            

80  =  (15/35)  - (7/35)N

80  =  (8/35) N

N = 80 (35/8)   =     350

originally, I think since n/h=1/3, h=3n, not n=3h

n-number of apples Nicole has at first

h-number of apples Hannah has at first

n/h=1/3

h=3n

(n+4)/(h+3)=1/2

2n+8=h+3

substitute h=3n

2n+8=3n+3

n=5

h=15

15+3=18

JP

Originally    n = 3 h

then     n+4  =  2 ( h+3 )       sub in the first equation

           3h + 4 = 2 ( h +3)

           h = 2              2 + 3 = 5 apples in the end for H

Just look at the he=ighest degree terms       x^4 - 2x^4  = - x^4        Degree remains  4  

please help me 

Angle C =  180    -  15  - 45 =    120

Using the Law of Sines

AC / sin B   =  AB /sin C

AC  / sin45  = 2sqrt 3 /  sin 120

AC =   2sqrt 3  * sin 45  / sqrt (3) / 2

AC  = 2sqrt (3)  (sqrt (2)  / 2 )  * 2 /sqrt (3)

AC =   2sqrt (2)

5x + 10  = 0

5(x + 2)  =  0

x + 2  =  0

x  =  -2

Using the Remainder Theorem

5(-2)^9   - 4(-2)^2  +  2   =

-2560  -  16   +   2   =

-2574    =  the remainder

Thr front and  back are composed of two    right triangles with legs of 3 and 4

Their combined surface area =   3 * 4  = 12

The  right side is a 3 x 8  rectangle =   24

The bottom is a  4 x 8  rectangle  =  32

The  top is a 5 x 8  rectangle = 40

Total surface area =   12  + 24 +  32  +  40    =    108  units^2

Bottom face is 8*4=32

2 triangle's area is 4*3*1/2*2=12

 5*8=40. 

back face is 8*3=24

32+12+40+24=108

-11+3i on the complex plane corresponds to (-11, 3) on the Cartesian plane, and 8+5i corresponds to (8, 5) through the same reasoning.  Because of this, the midpoint between the line segment connecting -11+5i and 8+5i is the midpoint of (-11, 3) and (8, 5) on the Cartesian plane, only turned into a complex number.  Thus, the midpoint is -1.5+4i

What I meant, for the first question, was that many other people have white space as well.

For the second question, even if there are seven lines, again, is it that big of a deal? 

For the third question, if you don't want to know the answer, why even look at the question then? As long as I have helped someone, and my answer is correct, I think that's good. If you simply ask nicely, and say "I think that too much white space isn't good" or something like that, that's fine. You can see that for some of my solutions (after you said that nobody likes too much white space), I completely understand. 

Also, I didn't mean to sound arrogant, supercilious, and self-aggrandizing. If I sounded that way (simply because I added seven white spaces, I'm sorry. Like I mentioned before, I don't use much white space (one line at most) anymore). 

there are 9 1-digit pages (9 digits)

there are 90 2-digit pages (180 digits)

there are 389 3-digit pages (1167 digits)

1167+180+9=1356 digits

JP

(r - 3)/sqrt(r) = -2*sqrt(r).         Multily both sides of the equation by  sqrt r

r-3  =  -2r

3r = 3

r = 1

By the Triangle Inequality, the maximum value is 2^6 = 64.

Do you realize that you have white space too?

Really? I’m so surprised you noticed.  I was using the Whiteness of Being communication protocol. It seemed appropriate, here....

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If there's literally four lines of white space, is it really that big of a deal?

For a mathematician, you do not count very well.  There are seven (7) lines of white space here. One of your other posts has one line of text with nine (9) white space lines. Some have none – such as on your more complex solutions. 

I don't know, but if I'm helping a person answer the problem, shouldn't you be grateful?

Perhaps I would be if I needed an answer or solution to the problem, but, in any case, I would be more grateful if I were not blinded by excess white space.

Three or four lines should be enough.  This electronic forum is not a sacred scroll nor is it a roll of toilet paper. There is no need to treat it as such. 

.

The number 7/2 + sqrt(15)/2*i cannot be a root.

The set of possible values of k is [1,5).

There are 685 subsets that contain the number 5.

The area is 3*sqrt(2).

Number of games in a season is x

Number of games you have played is 0.8x, so you have (1-0.8)x=(0.2)x

You have won (0.48)(0.8)x=0.384x. 

You must hae 0.5x wins, since you want the same number of wins and losses. So, you have to have 0.5-0.384=0.116x

(0.116x)/(0.2x)=0.58

58%

JP

https://web2.0calc.com/questions/combination-probability 

I don't know, but if I'm helping a person answer the problem, shouldn't you be grateful? If there's literally four lines of white space, is it really that big of a deal?

r = (2 + sqrt(3))/4, so a + b + c = 2 + 3 + 4 = 9.

I mean, if you don't like it, that's completely fine, just tell me. There's no need to be rude. Just saying. Also, look at your problem. Do you realize that you have white space too?