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Not a problem - I thought he might have.

I have already written quite a lot for this - they in the sticky topics already - there are actually 2 of them that need amalgamation.

Let me see.

Here is the newest one that I did yesterday

I asked CPhill, and he had recieved the message all along!

Sorry about that

It's only annoying when someone asks why it's annoying.

Ok, sounds good.

I'll try making a "rough draft" for the new information page, then post it for you guys or somthing.

If you guys have any sugestions or just hate how I aranged it, I can then edit it or delete it altogether. :)

As for the sticky topic thing, I can see what your guys saying, and I actually agree with you guys.

This forum has a very relaxed feel, and is probably why I liked it and joined!

Welcome, Strider.......I'll give you your first points for that nice post !!!

sometimes i wonder if these anons are just someone talking to themselves 

I believe the questioner is asking for the square root of 1/9, zegroes......

This = √(1/9) = √1/√9 = 1/3

0.111111111

1)   The edge length of the bigger cube is 9/8ths that of the smaller cube. To see this, note that the surface area of a cube is just 6s^2, where s is the side length.  So, let the edge length of the smaller cube be s and let the edge length of the larger cube be (9/8)s. So the ratio of the surface area of the larger cube to that of the smaller cube is......

[6*[(9/8)(s)]^2] / [6*s^2] = (9/8)^2 = 81/64

2) The ratio of the dimensions of the larger cylinder to the smaller one is just 5/3. Thus, the larger one has a radius 5/3rds times the smaller and a height 5/3rds times the smaller.  To see this note that the ratio of their volumes is just....

[pi*[(5/3)r]^2*(5/3)*h] / [pi*r^2*h] = (5/3)^3 = 125/27 .....just as we thought

So...it appears that the ratio of their volumes is just the cube of the scale factor (5/3). And volume is a "cubed" quantity. And since surface area is a "squared" quantity, the ratio of their surface areas is just (5/3)^2 = (25/9).

3)(a) If the scale was 1:40, the actual rocket was 40 times as long as the model...so 40 * 75cm = 3000cm. And since a meter = 100cm, we have 3000cm/100 = 30 m.

3)(b) We can do this one by something called "dimensional analysis"......Note that 1 square meter = (1m) *(1m)  = (100cm)*(100cm). Thus, a square meter = 10000cm^2  .......so we have

2750cm^2 * [(1m^2)/10000cm^2] = .275m^2 .....Notice how the cm. just "cancel out" leaving us with m^2  !!!

3)(c)  For this one, we just note that 1 cubic meter = (1m) *(1m)*(1m)  = (100cm)*(100cm)*(100cm). Thus, a cubic meter = 1000000cm^2  .......so doing the same sort of thing we did on the last one...

96m^3 * [1000000cm^3/1m^3] = 96000000cm^3.....again, note how the meters "cross cancel"

We have to use the inverse cosine to find this angle......thus

cos^-1(5/8) = 51.32°   ......Note that this could also be a 51.3° angle in the 4th quadrant since the cosine is positive there as well.  Thus, 360° - 51.3° = 308.7°

Well, you have to remember that this is just a 30° angle in the second quadrant. And the cos(30°) = √3/2.

But, since the cosine is negative in this quadrant, we have to take the negative value of this = -√3/2.

And that's it.....

i have a new issue now.

p = −2i − 3j and q =4i +7j. (Note: these vectors are 3D).

Does that mean there HAS to be a third? 

nevermind. i figured it out. two negatives is a positive. Thanks again for the help. 

sorry, think i did something wrong..my a × b= -4,11,4. 

so i'll call it y. y= a3 b1 - a1 b3. 

a = 3 − 2 + 5 and b = 7 + 4 − 8

y=(5)(7)-(3)(8)
 = 35-24
 =11

I noticed however you got 59..where did i go wrong? 

You're welcome.

thanks for the help

I used u and v as general vectors.  a and b are your specific vectors.  That is they are specific examples of u and v.  But if you prefer to think of it as turning a and b into u and v ok!  Whatever works for you!

In fact, why don't you rewrite my u and v in terms of a and b (including their components - so u1 becomes a1 etc.)?

Can every natural number be formed as the difference of two primes?

No!  Most odd numbers can't, unless they happen to be 2 less than a prime number, as the difference between two prime numbers is even, unless one of them is 2.

sorry, im a little confused. did you just turn a and b into u and v? 

I'm with reinout and zegroes on this.  The forum might be a little "messy" but a much more structured look/feel would probably put off several who need help but would not ask for fear of being laughed at by "serious" mathematicians.

Queue the Alice Cooper music for zegroes.........!!!

....No more pencils

No more books

No more teacher's

Dirty Looks.........

Suppose we have the quadratic....... ax^2 + bx + c

This technique will work as long as we can find another factorization of ac where the factors sum to "b"

To see why, let's suppose ac = n

Now, let us suppose that there are two other factors of n, say q and r, such that q + r = b....then...

 (a must equal q*m)  and (c must equal r/m)  because (qm) * (r/m) = q*r = n

So, we have..........

ax^2 + bx + c =

(q*m)x^2 + (q + r) x  + (r/m) =

(q*m)x^2 + qx  + rx + (r/m) =

qmx(x + 1/m) + r (x + 1/m) =

(qmx + r) ( x + 1/m)

Which shows that this type of polynomial can be factored as long as ac - i.e., n - has factors q,r such that q + r = b