All Answers 352717 Answers

Here is where it was asked last time.

Seems you will be accused of cheating....

First I will state that i haven ot attempted to work this out.

Someone else answered last time.

So you have worked out that these cannot be together

1,2,3

2,3,4

3,4,5

4,5,6

5,6,7

6,7,8

7,8,9,

8,9,1

9,1,2

Mmm

There are lots of others too.  like 3,6,9 etc

This is really tricky.

yeah cus i just tryed to change mine to check so either my computers acting up too or you cant

ok thanks :)

No i dont think you can change it

What have you done already towards solving this ?

Have you worked out which triples add to a multiple of 3?

This question was answered not that long ago.

I expect that you mean the numbers 1,2,3,4,5,6,7,8,and 9   

Is this what you mean?

\(log_3 x=81\\ 3^{log_3 x}=3^{81}\\ x=3^{81}\)

You go wrong on your fourth line.    Check:  2log(1/10) +2 =0     log(900) =  2.954

\(2log_{10}x+2=log_{10}900\\ 2log_{10}x+2=log_{10}30^2\\2log_{10}x+2=2log_{10}30\\ log_{10}x+1=log_{10}30\\log_{10}x+log_{10}10=log_{10}30\\log_{10}(10x)=log_{10}30\\ 10x=30\\x = 3\)

y+2x=5

and

y+2x=5

they are the same graph

so every point on the line   y=-2x+5     is a solution.

l+k+h=53

l=k+6

k=h+4

l=h+4+6=h+10

l+k+h=(h+10)+(h+4)+ h = 3h + 14

3h + 14 = 53

3h = 39

h=13

heidi is 13 (which means karen is 17 and louis is 23 )

:)

Hi Jurimagic,

It is an exponential graph.

5.5 to the power of anything must be bigger than 0

and so (5.5 ^x)+1 must be bigger than 1

As x get big in the positive direction  y gets VERY big.

When x=0 y=2    (edited)

As x tends to negative infinity, y gets closer to 1 (but it will always be bigger than 1)

so

y=1 is an asymptote.

Here is your graph

Try watching this

Try watching this

Hi Tom...riddle    

Thanks Nirvana   

I am going to say the same thing as Nirvana but  use slightly  different words.

(not better words, just different words)

There are 1200 students but you can think of this as 1200*5 faces in front of teachers.

So we have 6000 enthusiastic faces holding on to every word the teachers' say (in my dreams)

There are 30 faces in each class so there must be 6000 divided by 30 = 200classes

Each teacher teaches 4 of these so there are  200 divided by 4 = 50 teachers.

lol me on every test 

there's not a name for that specific thing, (it looks awesome btw) 

but theres something sort of similar 

pretend you are given a line segment and you are told to construct the perpendicular bisector 

what you'd do is to take your "compass" and make a bunch of concentric circles about 2/3 of the way past your mid point (on the line segment) 

and you would want to do this on both sides so from one end point to the other (and remember to keep compass setting the same!) 

and so these "concentric circles" wherever they intersect, you can draw a line from the top to the bottom and there you have your perp. bisector....

again, it's not the same it just kind of reminded me of that :') 

Find the minimum value of \(\dfrac{(x + y)^3}{x^2 y}\), where x and y are positive real numbers.

\(\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{\dfrac{(x + y)^3}{x^2 y} } \\\\ \dfrac{\partial f(x,y)} {\partial x} &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3(x+y)^3}{(x+y)^3} - \dfrac{2xy}{x^2y} \right) \\ &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{2}{x} \right) \\ 0 &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{2}{x} \right) \\ \dfrac{3}{x+y} - \dfrac{2}{x} &=& 0 \\ \dfrac{3}{x+y} &=& \dfrac{2}{x} \\ \dfrac{x+y}{3} &=& \dfrac{x}{2} \\ 2x+2y &=& 3x \\ \mathbf{x} &=& \mathbf{2y} \\\\ \dfrac{\partial f(x,y)} {\partial y} &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3(x+y)^3}{(x+y)^3} - \dfrac{x^2}{x^2y} \right) \\ &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{1}{y} \right) \\ 0 &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{1}{y} \right) \\ \dfrac{3}{x+y} - \dfrac{1}{y} &=& 0 \\ \dfrac{3}{x+y} &=& \dfrac{1}{y} \\ \dfrac{x+y}{3} &=& \dfrac{y} {1} \\ x+y &=& 3y \\ \mathbf{x} &=& \mathbf{2y} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{\dfrac{(x + y)^3}{x^2 y} } \quad | \quad \text{minimize } x=2y \\\\ f(x,y) &=& \dfrac{(2y + y)^3}{(2y)^2 y} \\\\ &=& \dfrac{(3y)^3}{4y^2y} \\\\ &=& \dfrac{27y^3}{4y^3} \quad | \quad y \text{ is a positive real number} \\\\ &=& \dfrac{27}{4} \\\\ \mathbf{f(x,y)_{\text{minimum}}} &=& \mathbf{6.75} \\ \hline \end{array} \)

Yes certainly Tom, 

Just try not to trivialise the answer for guest.

Most people learn nothing if they are not forced to think a little for themselves.

excuse my trash grammar btw im really braindead and tired :')

like they established already that 6000 comes from the 1200 students with the total of 5 classes among them....

soo,,,,,, and we originally know that in EACH class there is a total of 30 students so...... in order to find the ACTUAL number of classes 

 you had to take that 6000 and divide it by 30 bc you know that you have 6000 students and you also know that each class has thirty so...
 

30x = 6,000

think of it like that :D 



 

I’m surprised you know how to log on.

If you really cannot do the work, then you need to take easier courses. This will put you on an elementary level. 

If you just don’t give a sh it then some counseling may help.   Fucking with us is not an answer to your problems!