But only if you assume the first 3 equations are right and use diffrent axioms than those were use to! otherwise math is never wrong an the answer has to be 35 and the first 3 equations are simply wrong

If you divide 30 up into it's primefactors you'll get: 2,3,5 you have 3! options to choose so the answer is 6, if 1 wouldn't be missing. So the real answer is 7. (c)) BUUUT since 0 also is a whole number

It depends. If for example C is the tallest than A and than B, A would have to carry more than C and B almost nothing. You could also answer this question with: If there all the same hight than the one

I actually do have a problem with these types of pictures. On one site math is never wrong isn't it? So the two equations in the middle just have to be wrong, we don't have any prblems and therefore the

Es handelt sich hierbei wenn ich das richtig gesehen habe um ein Eulerkreis- bzw. Eulerweg-Problem. Vorraussetzung für einen Eulerkreis ist, dass alle Knotenpunkte eine gerade Anzahl an Verbindungen haben,

9

Depends. Could be 1, 54325 or 14325

90

96

Depending on which rule you apply there are diffrent solutions. I agree with the rest, that there is one (or more) rule (they probably are essentially the same thing) with which you'll get 30 as answer.

But I understood the task correctly? And if so is it ok to assume a 1:1 chance for boy:girl? Do we have to find the exact ratio we'd expect or is a proof which gender is more often enough?

10t+o=to+58 t(10-o)=58-o since t has to be single digit and o can't be more than 9: 49/9=5,... so 10-o has to be higher than 5 and therefore o lower than 5. if we start with o=4 54=6t t=9 and so o=4 and

x+17=2+2x 15=x 2x+2=32 The box is 32cm long

f=2/7t f+4=2/5(t-10) 2/5(t-10)-4=2/7t 1/5t-1/7t=4 2t=240 t=120 an therefor f=40 Which gives a total of 440$

Each couple can give birth to as many children as they want until they have one doughter? And the question is if there will be more girls or boys? Without having a proof by probability I'd say there are

Yes but squaring a number actually makes them lose their little sign which means minus. Therefore -0,5 and +0.5 are exactly the same after squaring them. So if you "remove" the square by taking

98

You'd have to calculate brackets before anything else! Taking the sqareroot of the square of something basically is the same as if you'd take the absolute value of the number and it doesn't matter if you

How about "incorrect math"?

time difference from the first to the second therm is 1/2t 2tx80+500=tx40+1100 120t=600 t=600/120=5 5x40+1100=1300 5x80=400 Marissa had $1300 and Janice $400