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#2 Help Me ! » Figuring out sin/cos/tan angel without calculator » 2006-09-11 13:57:55

Replies: 8

How do I do it?

Like sin 90 = 1 and sin 210 = -1

Is there a chart that list the values and the only way for me is to memorize them?

Or is there some kind of pattern I can follow?

#3 Help Me ! » Help with an optimization.. very wierd! » 2006-07-16 18:19:03

Replies: 3

A fuel tank is being designed to contain 200 m cube of gasoline; however, the maximum length tank that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle with hemispheres at each end. If the hemispheres are twice as expensive per unit area as the cylindrical wall, then find the radius and height of the cylindrical part so that the cost of manufacturing the tank will be minimal. Give the answer correct to the nearest centimetre

So I derive the relationship between R and H to this:

300 = (pie)(r²)(h)

But I dont know how to find the minimum price.

A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/ L of gas. Gas costs $0.68/L. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. Drivers are paid $35/h in wages and benefits. Fixed costs for running the truck are $15.50/h. If a trip of 450 km is planned, what speed will minimize operating expenses?

This one, I'm simply lost at the variables.

#4 Re: Help Me ! » Finding ways from A to B in 3d objects » 2006-06-18 16:56:19

Ahh... it seems I misinterpret the question wrongly, or I didnt word it correctly.

It was under my impression that they want to find the number of ways to reach A and B if you have to move along the lines. So for example, you can go RIGHT from A to the right most, DOWN, then UP.



#5 Help Me ! » Finding ways from A to B in 3d objects » 2006-06-18 13:38:32

Replies: 4

Find the total number of ways to go from A to B, if each way is the shortest route.


This is a rough sketch of 2 by 2 by 2 object.

I have to find the number of ways for x by x by x object and x by y by z object.

I tried doing it for 1 by 1 by 1 and 2 by 2 by 2 based on manual calcualting. However I cannot find a general formula for this.

I'm thinking x + y + z, but that's just a pure guess. I tried it on a paper, but it's too hard to show what I tried in this forum.

#6 Help Me ! » Countil problem » 2006-06-15 12:10:16

Replies: 6

The doughnut shop has 5 kinds of doughnuts: a, b, c,d and e. There are unlimited supply of each kind. In how many ways can you order a dozen doughnuts?

Well, my first instict is to simply 5^12. But then I realize aaaab is the same as baaaa... hence the order doesnt matter.

I'm trying to do it by cases:

Case 1: one daughnuts only.

(5 C 1) x 1 = 5 ways

Case 2: two daughnuts only.

(5 C 2) x 13 = 130
But this include one daughnuts, so its 130-5 = 125.

The problem starts here. I dont know why the arrangements for 2 daughnuts is 13. I got it by simply listing out all the cases.

Then I tried using simpler problems, like let's say you want to buy 5 daughnuts out of 3 different daughnuts. I can find out the number of ways for this one, but I see no relation if let's say you want to buy 6 daughnuts.

I sat down for 2 hours and still couldnt figure it out sad

My last bet is 13^4... but that's a wild guess. Help is appreciated.

#7 Re: Help Me ! » Arrangement problem » 2006-06-15 11:59:42

Thank you guys. The answer for no. 2 is all right.

However for the first one, the answer is 225. I'm not sure how.. I got 900 too.

#8 Help Me ! » Arrangement problem » 2006-06-14 18:59:51

Replies: 6

1. Ten trees - four pines, four cedars, and two spruce- are planted in two parallel rows of five trees. How many arrangements are possible if each row must have the same composition of trees, not necessarily in the same order.

2. Suppose we want to creat subsets of the ten digits {0, 1, 2, ...., 9}
  a) How many subsets can be created, including the empty set?
  b) How many of the subsets contain only digits less than 7?
  c) How many of the subsets contain 0 or 9?

#9 Re: Help Me ! » ... equation ... » 2006-05-31 21:29:51

3/x = 4/x

is the same as

x/3 = x/4

So x = 0 works in this case smile

Or you can multiple the quesiton by x² and you will get 3x = 4x, which gives you x = 0.

#10 Re: Help Me ! » ... equation ... » 2006-05-31 21:09:43

x = 0 counts as a solution right?

#11 Re: Help Me ! » Permutation » 2006-05-31 20:16:32

Thank you for explaining that! It makes sense now.

And 24 is the right answer ^^

Thanks once again!

edit : actually I dont really understand this part.

"25, 50, 75 have (5x5) as their factor, which in effect means, they are made of two 5s. Hence, add 3 more zeros. Although 100 also as (5x5) as its factor, remember, we have already taken the two zeros of 100 into account"

Dont you already take 25, 50, 75 into account?

#12 Re: Help Me ! » Permutation » 2006-05-31 19:34:49

Calculator can only do up to 1 x 10^99. Any more of that and it will simply overflow.

100! is an overflow

10^100 is an overlow too.

#13 Re: Help Me ! » Permutation » 2006-05-31 19:23:59

I thought of that too, but calculator is useless since they will give you "overflow" message.

I'm trying to solve it by paper... but no luck so far.

edit : the 100! is actually 100 factorial. The ! wasnt the ending of the question heh.. it kinda mislead people.

#14 Help Me ! » Permutation » 2006-05-31 18:57:54

Replies: 19

Find the largest value of k so that 10^k divides evenly into 100!

To be honest, I dont really understand what the quesiton is asking. Help is appreciated!

#16 Re: Help Me ! » Proving with vectors » 2006-04-30 10:19:45

Thanks Ricky... but I need to prove it with vectors method. Either position vector or point-to-point vector.

Troublesome I know sad

#17 Help Me ! » Proving with vectors » 2006-04-30 07:56:07

Replies: 3

1) Prove that the median to the base of an isosceles triangle is perpendicular to the base.

2) In triangle ABC, the points D, E, and F are the midpoints of sides BC, CA, and AB, respectively. The perpendicular at E to AC meet the perpendicular at F to AB at the point Q.

a. Prove that AB dot (QD - ½AC) = 0

b. Prove that AC dot (QD - ½AB) = 0

c. Use parts a and b to prove that CB dot QD = 0

d. Explain why these results prove that the perpendicular bisector of the sides of a triangle meet at common point.

Thank you in advance!!

#18 Re: Help Me ! » Similar triangles » 2006-02-28 18:14:48

I tried doing it that way. Let's say x = triangle ABC

ADE / ABC = 3 / 7

81/x = 3/7

x = 189

But the answer is 441

#19 Re: Help Me ! » derivatives » 2006-02-28 17:56:56

I'm positive that it's asking for an inverse. But I dont know why you are dealing with inverse when you have learnt derivatives.

#20 Re: Help Me ! » Similar triangles » 2006-02-28 15:55:46

oh those are the areas I mean.

#21 Help Me ! » Similar triangles » 2006-02-28 15:30:27

Replies: 6

This question is really easy but I'm weak in similar triangle.. I dont know how to do it.

In triangle ABC, D and E are in AB and AC respectively such that DE is parallel to BC, AD = 3, DB = 4, and triangle ADE = 81. Determine

1. triangle ABC

2. quadrilateral DBCE

#22 Re: Help Me ! » Quadratics Again.. » 2006-02-26 13:22:25

That's true. You can draw your quadraic anyhow and you can still make it pass through those 2 points. A straight line, a third degree function, a fourth degree function etc..

#23 Help Me ! » Need help for a triangle » 2006-02-23 21:57:18

Replies: 2

ABCD is a trapezoid in which AB is parallel to DC and AB = 9, DC = 15. The perpendicular distance between AB and CD is 4. Sides DA and CB are extended and meet at point E. If F and G are midpoints of Da and CB respectively, determine the area of triangle FGE.

#24 Help Me ! » Discreet proof help needed » 2006-02-21 20:12:41

Replies: 4

1) Prove that no square integer number can have a remainder of 3 when divided by 5

2) A school has 1000 students, and their lockers, which are numbered from 1 to 1000, are all closed, The first student opens all the lockers. The second student closes every second locker, beginning with her locker #2. The third student CHANGES the state of every third locker, beginning with locker #3, which means a locked locker becomes open and an open locker becomes closed. This carries on until all 1000 students had their turn.

Which lockers are open and why?

After some work, I think the locker with perfect square numbers are open. So locker #1, 4, 9, 16, 25 etc are open. But I dont know how to prove this and it might not work for numbers within 900-1000.

Also, I need help solving the first problem in this thread :

#25 Re: Help Me ! » Geomtery Proof » 2006-02-19 22:34:58

Thanks for the solution to the second problem. Your coloring makes it so easy to understand.
Thanks once again ^^

Actually I have one more problem I'm stuck with, but I'm satisfied with only these 2 solutions ^_^

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