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I put 337.5% as my answer. It is incorrect.

Last weekend, Amy raked leaves in her front yard. She noticed that 11/3 pounds of leaves filled 4/9 of a bag. What is the unit rate for pounds of leaves per bag?     

Here's a little tip:  per means divided by. 

When you see the word "per" substitute "divided by"

so lbs per bag means 11/3 divided by 4/9 in this problem. 

_.

For the hour hand :

  Each hour (12 of them) covers 30 degrees     (360/12 = 30 degrees)

    at 2:00  you have covered 60 degrees

      Each minute the hour hand moves  1/2 degree    (30 degrees / hour  x  1 hour/60 min = 1/2 degree)  

          so at 2 48   the hour hand is at    2*30  + 48 * 1/2 =84 degrees

For the minute hand:

  Each minute is 6 degrees     (360 degrees   x  1 hr/ 60 min =  6 degrees / min)

    so at 48 minutes the minute hand is at    48 x 6 = 288 degrees

288 - 84 = 204 degrees  

*******

You are given   LB   and  BAG     and you want   LB/BAG

11/3 LB  /  4/9 BAG  = 11/3  * 9/4  = 99/12  LB/BAG = 8 1/4  LB / BAG

Thanks Ginger.    

When x = y = 1/2, x^4*y + xy^4 = 1/16, so this is the maximum value.

He has already done it for all his questions Ginger.

ok thanks.

Maybe you are right.

In which case I offer my appologies to ZZZZZZZZZZ.

Sorry ZZZZZZ

He could have just sent the whole class the address of it.

It is easy to do that with Desmos.

I think that this is what zzz had in mind.

.\(\displaystyle \text{Given that }a^{2}-3a+9=0,\dots(1),\\a^{2}=3a-9,\\\text{so}\\a^{3}=3a^{2}-9a.\\ \text{But from (1), }\\a^{2}-3a=-9,\\ \text{so}\\3(a^{2}-3a)=a^{3}=-27.\)

Answering guest has given you an extremely good outline. Which is very similar to what i did.

Use it to your advantage.

Please no one answer more without good interaction from asker guest first.

Note:

I am not saying my final answer was necessarily correct. I did it in a hurry.

Hi EP

Do you feel the need to do this rude guest's homework for him?

If he interacts again politely that would be a bit different, then it would make sense to help him a bit.

Hi CalculatorUser and guest.

Guest does know why he thinks it is wrong.

Most likely because his book or his teacher says it is wrong. 

If he has the answer that he believes is correct he should provide it.  (Even if it just a number with no explanation)

Guest: It is extremely rude just to put the word 'wrong'

You need to say why and you should thanks the answerer for their time and interest in your question.

Withdrawn.

I understand the need to not answer but I am interested in this problem too.

I do not know the answer.

What is wrong with me discussing it here?

I know you are going to tell me it is against the rules.

Is it against the rules for students to discuss it with each other?

What is the difference, I, like the other students, do not know the answer.

Picture the horizontal equilateral triangle formed by the centres of the three 'base' spheres.

The centre of the fourth sphere will lie directly above the centre of this triangle forming a tetrahedron with the other three centres.

The lengths of the sides of the base will be 2, and the sloping sides 3.

Now calculate the height of the tetrahedron and add 3.

Very welcolme for what ?

A largely useless and irrelevant answer ?

You have not answered the question.

You begin with an approximation, the best that you can hope for is an approximation to the exact result.

All that you have succeeded in showing, with the aid of a calculator,  is that the value of the expression is somewhere in the neighbourhood of 1.

You don't seem to have much of an idea of what mathematics is about.

When looking at a problem like this, the first thing that you should do is to ask whether there is something significant about the numbers chosen

or whether any old set of coefficients/numbers would have done.

If you had stopped to do that you might have seen the significance of the coefficients 1 4 6 4.

\(\displaystyle x^{4}+4x^{3}+6x^{2}+4x-7\\=x^{4}+4x^{3}+6x^{2}+4x+1-8\\ =(x+1)^{4}-8\\= (\sqrt{3}-1+1)^{4}-8\\=9-8=1.\)

it's okay Melody! you always help me anyway :) and I really appreciate it 

thank you! :) 

You're really not supposed to be discussing this yet... people could still be taking it until the 19th. I know all the schools in my area will have to take it at a date later than the 12th, which might be Monday or Tuesday.

AoPS.Morrisville PLEASE do not post homework questions on this forum

Sorry,I posted an EXCEL chart without explanation. For the \(2 \times 50\) grid, the number of arrangements is

\({50 \choose 0}+{49 \choose 1} + {48 \choose 2} + ...+{26 \choose 24} +{25 \choose 25}= 20,365,011,074\) arrangements. EXCEL was the only tool I could use to calculate such an enormous sum.

I am looking for a more elegant way to count this giant number of arrangement. I am certain there is one somewhere on the web!

Since the ellipse is centered at the origin and its foci are on the y axis and it passes through the point (-4,0), then the minor axis must  be  8  units long  and the major axis must be 24 units long

So.....b = 8/2  = 4

And a = 12/2 = 6

So.....the equation of the ellipse is

x^2          y^2

___  +   _____    = 1

4^2          6^2

x^2        y^2

___ +   ____   =   1

16           36

Here's a graph :   https://www.desmos.com/calculator/qdv0dposwj