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first try: 0.70 * 20 = 14 passed , 6 failed
second try: 0.5 * 6 = 3 passed
so 14 + 3 = 17 out of 20 passed
100(17/20) = 5*17 = 85%

What do you don't understand?

Wait, I find something that would be right, am I?

\((184-120)*\frac{1}{5}\) and this is: (\(12.8\)), isn't it like 128 balloons? (128 balloons for Mel that he got at the end)

And if you got 128 ballons, add 184 balloons, and then you got (\(128+184=\)) 312 balloons, 

and the 312 balloons are for Luis. ​Suddenly just remembered that, the answer.

​​

Now that's a explanation, or not?

Straight

70% passed the test: (\(\frac{20}{100}*70\)) is 14.0,

so 14 students passed at the first try. (20 students)

30% failed the test: (\(\frac{20}{100}*30\)) is 6.0,

so 6 students failed at the first try. (20 students)

The half of 30% is (\(30:2=\)) 15%.

Imagine if the 30% would be 100%, means all students failed, 

and at total they're only 6 students, so wouldn't it be 15%,

but it would be the half of 100%, so: (\(100:2=\)) 50%.

6 students are 100% and 50% would be (\(6:2=\)) 3 students.

At the second try 3 students will pass and the other 3 stud. will fail.

This make no sense in LaTeX or calc.

Do not write something that does not make any sense.

If n would be 5 (\(n=5\)), then would be the onliest possibiity for m that is positive (\(m=8\)).

Note that I already say 25 for \(n^2\)

\((m+n)^2+(m-n)^2+m^2+n^2+51=318\)

\(m^2+10m+25+m^2-10m+25+m^2+25+51=318\)

\(3m^2+126=318\)

\(3m^2=318-126\)

\(3m^2=192\)

\(m^2=64\)

\(m=8\)

Straight

Luis has (\(280\)) balloons and Mel has (\(160\)) balloons, difference: (\(120 \)).

Mel gives him (\(\frac{1}{5}\)) balloons, so that means he gives him (\(32\)) balloons.

Now Luis has (\(312\)) balloons and Mel has (\(128 \)) balloons.

Straight

Thank you so much

a) 24 possibilities, because \(4!\) is 24 (\(4 * 3 * 2 * 1\)).

b) There were 4 paper squares and they are cutting every paper square along a diagonal (1 paper square = 2 triangles), so 4 paper squares are 8 triangles (\(4 * 2\)). We're calling every triangle a, b, c, d, but a (b, c and d) has 2 triangles (a and a, b and b, c and c, d and d). With a you can make a paper square with: a, b, c and d, so: aa, ab, ac, ad or aa, ba, ca, da. The same is with b, c and d (but note that squares are considered equal if they fall apart due to rotations and shifts!) : bb, bc, bd or bb, cb, db. For c: cc, cd or cc, dc. And at the end for d: dd. If you count them together there are 10 possibilities to make paper squares.

The question said: How many of the squares are two-coloured? It only can be under 10 possibilities (\(n<10\)). It applies that aa, bb, cc and dd couldn't be two-coloured, because they're having the same colour. Four cannot be possible, so there are (\(10 - 4=\)) 6 possibilities, namely: ab, ac, ad, bc, bd and cd.

c) Like b) said that there are 6 paper squares two-coloured and he wanted to choose two of them. Let us call: ab as 1, ac as 2, ad as 3, bc as 4, bd as 5 and cd as 6. (Note: Squares are considered equal if they fall apart due to rotations and shifts.) He can choose: 12, 13, 14, 15,16; 23, 24, 25, 26; 34, 35, 36; 45, 46; 56. If you count them there are 15 possibilities. There are 15 ways.

Am I right or not, Melody?

Straight

So tell us what you have done.

Did all, but I am not sure.

I assume the function is sin(x)*sin(x+pi/3):

asinus is right.  I did the last bit wrong.

x= pi/3      or  x = 5pi/3

5!=120

so it looks like you want the middle one.

so it will start with a 3  

How many are less than 30000?    That would be 2*4!=48

how many start with 31?    That would be 3! = 6

So another 6 will start with 32

So that means there are 60 that are less than 34000

So the 60th one must be  32541

cos(2x)-5cos(x)+3=0

이 식에서 x의 해를 구해야 합니다.

Hello Guest!

\(cos(2x)=2cos^2(x)-1\\ 2cos^2(x)-1-5cos(x)+3=0\\ cos^2(x)-2.5cos(x)+1=0\\ cos(x)=1.25\pm\sqrt{1.25^2-1}\\ cos(x)\in \{0.5,2(not\ applicable )\}\)

\(x=acos(0.5)\)

\(x\in \{60°,300°\}\)

  !

Have you tried doing any of this for yourself?

cos(2x)-5cos(x)+3=0

\(cos(2x)-5cos(x)+3=0\\ cos^2x-sin^2x-5cosx+3=0\\ cos^2x-(1-cos^2x)-5cosx+3=0\\ 2cos^2x-1-5cosx+3=0\\ 2cos^2x-5cosx+2=0\\ let\;\;y=cosx\\ 2y^2-5y+2=0\\ y=\frac{5\pm\sqrt{25-16}}{4}\\ y=\frac{5\pm 3}{4}\\ y=\frac{8}{4}\qquad y=\frac{2}{4}\\ cosx=2\qquad or \qquad cosx=0.5\\ cosx=0.5\\ x=2\pi n \pm \frac{\pi}{6}\qquad n\in Z\\ x=\frac{\pi}{6}\quad or \quad x=\frac{11\pi}{6}\qquad 0

The LaTex is not rendering.   I have taken a pic of what it is supposed to be looking like.

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Cathy spent 3/4​ of her pocket money. If she spent 1/3​ of it on a movie, 1/6​ on books, and the rest on a dress, what part of her pocket money did she spend on the dress?

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Tricky question...

Let's write down some numbers: 

Cathy's pocket money = $100

Money she spent         = $75

Movie                           = $25

Books                           = $12.50

Dress                           = $37.50

The part of the pocket money she used to buy a dress = 3/8