Melody

Hi  Max,

So how did you do it?

I see lots of irrelevant questions and they do not activate the flag ://

1/2(x+1)+1/3(x+4)=1/6(x-1)

This is the question you have asked

\(\frac{(x+1)}{2}+\frac{(x+4)}{3}=\frac{(x-1)}{6}\)

The easiest way to solve it is to multiply both sides by the smallest common denominator, which is 6.

That will get rid of all the fractions for you.

Give it a go and if you want more help post what you have done.

.How many ways can 11 Redcoat basketball players and Coach Woodburn (12 people) be arranged along the sidelines if Travis and Pat must stand together?

More info needed.

0ne sideline or two side lines.  Do equal numbers of people need to be on each sideline? Maybe there are other considerations too but I haven't thought of them yet.

-9/58=(2y-(3y+4))/(7y-(2-5y))

\(\frac{-9}{58}=\frac{2y\;\;-1(3y+4)}{7y\;\;-1(2-5y)}\\ \frac{-9}{58}=\frac{2y\;\;-3y-4}{7y\;\;-2+5y}\\ \frac{-9}{58}=\frac{-y\;\;-4}{12y\;\;-2}\\ \frac{-9}{58}=\frac{-(y\;\;+4)}{2(6y\;\;-1)}\qquad \mbox{note that }6y-1\ne0\quad so\;\;\;\;y\ne \frac{1}{6}\\ -9*2(6y-1)=58*-1*(y+4)\\ 9*2(6y-1)=58*(y+4)\\ 9(6y-1)=29(y+4)\\ 54y-9=29y+116\\ 25y=125\\ y=5\)

-9/58=(2y-(3y+4))/(7y-(2-5y))

check:

(2*5-(3*5+4))/(7*5-(2-5*5)) = -0.1551724137931034 = -9/58

RHS=(2*6-(3*6+4))/(7*6-(2-5*6) = -0.1428571428571429​

LHS=-9/58= -0.1551724137931034

a= sqrt3(216 )

\(a= \sqrt[3]{216 }=6\)

I really do not know what you (your teacher) is asking about logs for on this particular question

Maybe

\(216^{1/3}=a\\ so\\ log216^{1/3}=loga\\ log(6^3)^{1/3}=loga\\ \frac{log(6^3)}{3}=loga\\ log(6^3)=3loga\\ log(6^3)=log(a^3)\\ log(6^3)=log(a^3)\\ so\\ a=6 \)

But this whole rigmarole is quite silly.

Presione [2nd] entonces [acos]

C

Well they look like tangents but if that is not known for sure then I guess the answer is C

e^(sqrt(-1)pi)+1

Here is the maths for it :)

\(e^{\sqrt{-1} \pi }+1\\ =e^{i \pi }+1\\ =Cos\pi+isin\pi+1\\ =-1+i*0+1\\ =-1+1\\ =0\)

Hey, guest, since you think "Maths is the Best" why don't you join up.  We want people like you to be a recognised part of our community :)

20x5/30+10x4/4x20/35

\(\frac{20\times5}{30}+\frac{\frac{10\times4}{4}\times20}{35}\\ =\frac{20\times5}{30}+\frac{\frac{10\times\not{4}}{\not{4}}\times20}{35}\\ =\frac{20\times5}{30}+\frac{10\times20}{35}\\\)

You should be able to take it from there. :)

It is interesting when these types of contradictions spring up :)

Thanks Alan and Michael  :)