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Here is my naive approach:
Subsets [O, O, O, O, F, F, F, T, T, H, {3} ] =53
 O=$1, F=$5, T=$20 and H=$100
{F, F, F} | {F, F, H} | {F, F, O} | {F, F, T} | {F, H, F} | {F, H, O} | {F, H, T} | {F, O, F} | {F, O, H} | {F, O, O} | {F, O, T} | {F, T, F} | {F, T, H} | {F, T, O} | {F, T, T} | {H, F, F} | {H, F, O} | {H, F, T} | {H, O, F} | {H, O, O} | {H, O, T} | {H, T, F} | {H, T, O} | {H, T, T} | {O, F, F} | {O, F, H} | {O, F, O} | {O, F, T} | {O, H, F} | {O, H, O} | {O, H, T} | {O, O, F} | {O, O, H} | {O, O, O} | {O, O, T} | {O, T, F} | {O, T, H} | {O, T, O} | {O, T, T} | {T, F, F} | {T, F, H} | {T, F, O} | {T, F, T} | {T, H, F} | {T, H, O} | {T, H, T} | {T, O, F} | {T, O, H} | {T, O, O} | {T, O, T} | {T, T, F} | {T, T, H} | {T, T, O} 

HAHAHA!!!....yeah....that can happen   !!!!

LOL!!!!!

Lightning solved it by NOT solving it......!!!!!

Which brings to mind the old paradox...."If every man in the village who is not the barber shaves himself.....who shaves the barber ? "

Thanks, CPhill. Some of these tick me off, because looking at how you did it, it was the same way I did, I just took one tiny wrong step and got WAYY off base.

Yes, you are absolutely correct.  Great observation!

My last sentence is completely wrong.  (I willl delete it with an explanatory note)

I think you can still use the beginning of my explanation to determine the correct answer.  I may do it myself when I have a little more time.

You didn't solve anything, Wolfram|alpha did.

Note that   sin^3  θ + cos^3  θ  factors as a sum of cubes   =  ( sin  θ  + cos  θ) ( sin^2 θ - sin θ cos  θ  + cos^2  θ)

So....the first  factor  cancels with the factor in the denominator  on the right and we are left with

(sin^2  θ -  sin θ cos  θ + cos^2  θ)  =

( sin^2  θ  + cos ^2  θ)  -  sin θcos θ  =

1     -  (1/2)sin (2 θ)

Great reminder at the bottom there.

tan (v/2)  =

sin(v/2)             √  [ (1 - cosv) / 2 ]             √ [ 1 - cos v]   

_______   =       _______________  =    _____________   multiply   top/bottom by √ [1 - cos v ]  and we have

cos(v/2)            √ [ (1 + cos v)/ 2 ]              √ [ 1 + cos v ]

√[1 - cosv ] √ [ 1 - cos v ]            1  -  cos v                 1  - cos v               1  - cos v

____________________ =       ____________  =     ___________  =   _________  =

√ [ 1 + cos v]√ [ 1 - cos v]         √ [ 1 - cos^2v ]            √ sin^2v                   sin v

 1         -      cos v

____           _____    =

sin v            sin v

csc v  -   cot v

This problem was answered before

There are 2^10 different subsets in the original set.

Now let us just subtract subsets without a 5

Let us just pretend you have set without a 5

So {1 2 3 4 6 7 8 9 10} Is your new set

So how many subsets can you make from that?

2^9 subsets

2^10 - 2^9 = 512 different subsets

NOTE: 2^10 - 2^9 is NOT 2.

So there you have it! :D:D:D

6. A spinner is divided into 5 sectors as shown. Each of the central angles of the sectors 1 through 3 measures 60° while each of the central angles of sectors 4 and 5 measures 90°. If the spinner is spun twice, what is the probability that at least one spin lands on an even number? Express your answer as a common fraction.

P(even on one at least of the spins) =   1 - P(odd on both spins)

P(odd on 1st spin)  =  [60 + 60 + 90] / 360  =  210/360  = 7/12

And we have the same probability of an odd on the second spin

So

P(odd on both spins)  =  7/12 * 7/12  =   49/144

So....P (even on at least one of the spins)  =  1 - 49/144  =  95 / 144

1. 3!=6 ways to arrange the three-digit number...and 132, 312 are multiples of 4. Thus, 2/6=1/3?

Solved it! Plugged it into Wolframalpha and got 3!

Nice, hectictar   !!!!

This is only for the second question:

We can multiply each dimension by  4  to make each dimension an integer. This will only scale the prism and the cubes up by  4....it won't change the number of cubes that can be fit into the prism.

So the new dimensions of the prism are  18 ft.  ×  30 ft.  ×  45 ft.

We can't cut the prism into cubes with sides length 18 ft, because 18 does not fit evenly into  30.

We can't cut the prism into cubes with sides length  9  ft, because  9  does not fit evenly into  30.

We can't cut the prism into cubes with sides length  2  ft, because  2  does not fit evenly into  45.

The side length of the cubes must fit evenly into 18, 30, and 45.

That is, the side length of the cubes must be a factor of 18, 30, and 45.

The maximum side length of the cubes is the greatest common factor of 18, 30, and 45.

The maximum side length of the cubes is  3 (ft.)

After splitting the prism into cubes with sides length 3 ft,

the prism will be 6 cubes high, 10 cubes wide, and 15 cubes deep.

the total number of cubes  =  6 * 10 * 15  =  900

Melody why is that? we can divide 2012! by 2 more than once

This is the third time you have asked this exact same question

Do not multiple post like this!

It is not urgent at all but here is the answer

Find the largest prime number p such that when 2012! is written in base p, it has at least p trailing zeroes.

If you write any number in a different base then the number of zeros will be the number of times that base can be divided into the number

For example.   20 written in base 2 will have 2 zeros at the end because 20 can be divided by 2 twice and get an integer answer.

I have deleted what I wrote after this point because it was nonsense.  BUT

You should be able to use this initial fact to help determine the correct answer.

THX, Rom  !!!

Here's the c part

Using implicit differentiation....the slope of a tangent line at any point on the ellipse is given by

(1/4)x + (1/2)y y'  = 0

(1/2)yy' = -(1/4)x

y' =  -(1/4)x / (1/2)y  =  -x / [ 2y]

And the slope of a tangent line to the hyperbola at any point is

2x - (2/3)yy' = 0

-(2/3)yy' = -2x

y' = 2x /[ (2/3)y ] =   3x / y

To find the itersection points.....we have that

(1/8)x^2 + (1/4)y^2  = 1    [multiply through by - 8 ] ⇒ - x^2 - 2y^2  = -8      (1)

x^2 - (1/3)y^2   = 1     (2)         add (1)  and (2) and we have that

-(7/3)y^2 = -7    [divide through by -7 ]

(1/3)y^2  = 1

y^2  = ±3

y = ±sqrt(3)

And using  (1)

-x^2 - 2(3) = -8

-x^2 - 6 = -8

x^2 + 6 = 8

x^2 = 2

x =±sqrt (2)

So.....the intersection points are

(sqrt(2), sqrt(3) )

(- sqrt(2),sqrt (3) )

(-sqrt(2), -sqrt(3) )

(sqrt (2), - sqrt(3))      

Testing the first point 

The slope of the tagent line to the ellipse at (sqrt(2), sqrt(3)  )   we have     -sqrt(2) / [2 sqrt(3)] = -1 / sqrt(6)

The slope of the tangent line to the hyperbola at this point is  3sqrt(2) / sqrt(3) = sqrt(6)

Since these are negative reciprocals.....the tangent lines are perpendicular

[I'll let you test the other three points ]