Math Problems!........share It Here!

[twisted minds]

got a problem from a text message. pasensya nya englishin ko lang....

 

there are 20 floors in the bldg. you have to pay half of your money on hand per each floor and the gaurd will give you a change of P1. (piso).

 

1. how much you need you have (minimum) inorder to reach 20th floor and still you have 2pesos on hand.

2. what is the algebraic formula or equation to solve this problem?

 

try ko lang using series. i can't type using the sum symbol so please bear with me ...

 

If T represents the amount you bigin with, and Xn represents the amount remaining at a certain floor, we get:

 

X0 = T

X1 = T/2 + 1

X2 = (T/2 + 1)/2 + 1 = T/4 + 1/2 + 1

X3 = (T/4 + 1/2 + 1)/2 + 1 = T/8 + 1/4 + 1/2 + 1

X4 = (T/8 + 1/4 + 1/2 + 1)/2 + 1 = T/16 + 1/8 + 1/4 + 1/2 + 1

.

.

.

Xn = T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) + ... + 1/(2^(n-n))

 

= T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) ... + 1

 

= (T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n)

 

Now, if R represents the remainder after the nth iteration (floor), we can get T as:

 

(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n) = R

 

(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) = R*2^n

 

T = R*2^n - 2 - 4 - 8 - ... - 2^(n-1) - 2^n

 

 

So suppose R = 2 after n = 20:

 

T = 2*2^n - 2 - 4 - 8 - ... - 2^(20-1) - 2^20

 

= 2^21 - 2 - 4 - 8 - ... - 2^19 - 2^20

 

= 2

 

Funny thing about this formula: as n approaches infinity, no matter how much you start with, Xn also approaches 2

[The_Blade]

Let X = money mo sa top floor (in this case 20th)Let N = number of floors below the top floor (in this case if n=1, youre at the 19th floor)Total amount of cash at floor N below the top = (2^N)*X + 2 - 2^(N+1) --> try it. Eto yung pera mo bago mo bayaran yung guard.So bale kung 2pesos lang matira sa yo at the top, X = 2. Plugging that into our equation:(2^N)*2 + 2 - 2^(N+1) = Cash |this will become (2^(N+1))2^(N+1) + 2 - 2 ^(N+1) = Cash //yung dalawang 2^(n+1) will cancel out2 = Cashthus showing that if the money left at the top floor = 2 pesos, the money at any floor below the top floor is equal to 2 pesos.

just want to say thanks!

try ko lang using series. i can't type using the sum symbol so please bear with me ...If T represents the amount you bigin with, and Xn represents the amount remaining at a certain floor, we get:X0 = TX1 = T/2 + 1X2 = (T/2 + 1)/2 + 1 = T/4 + 1/2 + 1X3 = (T/4 + 1/2 + 1)/2 + 1 = T/8 + 1/4 + 1/2 + 1X4 = (T/8 + 1/4 + 1/2 + 1)/2 + 1 = T/16 + 1/8 + 1/4 + 1/2 + 1 . . .Xn = T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) + ... + 1/(2^(n-n)) = T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) ... + 1 = (T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n)Now, if R represents the remainder after the nth iteration (floor), we can get T as:(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n) = R(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) = R*2^nT = R*2^n - 2 - 4 - 8 - ... - 2^(n-1) - 2^nSo suppose R = 2 after n = 20:T = 2*2^n - 2 - 4 - 8 - ... - 2^(20-1) - 2^20 = 2^21 - 2 - 4 - 8 - ... - 2^19 - 2^20 = 2Funny thing about this formula: as n approaches infinity, no matter how much you start with, Xn also approaches 2

thanks din...

[FORUMsubscriber]

m log [ F/P - 1 ] raise to n times 100

[FORUMsubscriber]

Is this the one from the calculator?

 

If T represents the amount you bigin with, and Xn represents the amount remaining at a certain floor, we get:

 

X0 = T

X1 = T/2 + 1

X2 = (T/2 + 1)/2 + 1 = T/4 + 1/2 + 1

X3 = (T/4 + 1/2 + 1)/2 + 1 = T/8 + 1/4 + 1/2 + 1

X4 = (T/8 + 1/4 + 1/2 + 1)/2 + 1 = T/16 + 1/8 + 1/4 + 1/2 + 1

.

.

.

Xn = T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) + ... + 1/(2^(n-n))

 

= T/(2^n) + 1/(2^(n-1)) + 1/(2^(n-2)) + 1/(2^(n-3)) ... + 1

 

= (T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n)

 

Now, if R represents the remainder after the nth iteration (floor), we can get T as:

 

(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) / (2^n) = R

 

(T + 2 + 4 + 8 + ... + 2^(n-1) + 2^n) = R*2^n

 

T = R*2^n - 2 - 4 - 8 - ... - 2^(n-1) - 2^n

So suppose R = 2 after n = 20:

 

T = 2*2^n - 2 - 4 - 8 - ... - 2^(20-1) - 2^20

 

= 2^21 - 2 - 4 - 8 - ... - 2^19 - 2^20

 

= 2

 

Funny thing about this formula: as n approaches infinity, no matter how much you start with, Xn also approaches 2

[destron]

A quite simple question:

 

The area of a rectangle is the product of its length and width, or;

 

A=LW

 

How is it derived?

 

 

i wish they could make a common sense theorem in geometry... anyhoo... basically derived from a square. a square is a rectangle with all sides equal and at right angles. the area for a square is basically side squared. so start with a square with an area of one. now figure in a rectangle and any given length or width. start with an example of rectangle with one side hvaing the measure of 1 for width and a measure of 4 for length. so what we do is place those 4 squares in a line... so thats a rectangle with a width of 1 and a length of 4. simply add up the areas of those four squares with an area of 1 and you get 4. so how do you derive 4 from the measure of side with a measure of 1 and a measure of 4? well logice precludes that we can follow the basic measure used by deriving the area of the square which is side squared... or a change in linguistics... side times side. so 1w times 4l equals the area of the 4 squares forming the rectangle.

 

but honestly... this really does not need any explanation at all. seriously.

 

Yep. No need.

Gee my signal on calculus is now just 1bar, so forgive me if I post erroneously:

 

 

The area of a surface (can be a rectangle) is:

 

∫f(x)dx

where the limit is 0 to W(width).

 

But since f(x) is a constant L(length):

 

A = ∫Ldx (lim 0 to W)

 

= L∫dx

 

=L(W - 0)

 

A = LW

[destron]

A quite simple question:

 

The area of a rectangle is the product of its length and width, or;

 

A=LW

 

How is it derived?

 

I said before that that was a simple question because my real question is just a tad more complicated, so I just decided to ask that one.^

 

Here goes my intended question:

 

The circumference of a circle is 2 pi times the radius, or;

 

C=2πr

 

How is it derived?

Edited October 21, 2007 by destron

[SonBorj]

2 pi R is the integral of the area of a circle which is pi R squared as R changes.

[gineh]

I said before that that was a simple question because my real question is just a tad more complicated, so I just decided to ask that one.^

 

Here goes my intended question:

 

The circumference of a circle is 2 pi times the radius, or;

 

C=2πr

 

How is it derived?

 

Sige, subukan ko. Medyo matyagaan ito.

 

We start by computing the perimeter of a square. But we wont be doing it by simply adding the sides.

 

http://img147.imageshack.us/img147/9320/squarewi5.jpg

Let's compute the perimeter of this square in terms of r only. Our attack shall begin from triangle at the bottom right with angles phi, theta and a right angle and hypotenuse r.We note the following first.

Theta = 360degrees / 8.

Imagine that the square was sliced into 8 pizza slices. R would be the length of one side and theta would be the

angle of the slice's "tusok". Theta's value would then be equal to 360degrees/8.

 

We can then compute the value of phi. We know that the sum of angles in a triangle is 180degrees. We know that the value of theta is (360/8). We also know that the other angle in the triangle is a right angle (90 degrees). We can then compute the value of phi.

phi = 180 - 90 - (360/8).

phi = 90 - (360/8)

We now know the value of phi. Let us then derive the value of a. We know that r is the hypotenuse and phi is the adjacent angle to a. Using trigonometry.

a = r x [cos(phi)]

And finally we compute the perimeter of the square. Let us notice that the square's perimeter is composed of eight As as each side of the square is composed of two As.

Perimeter = 8 x a

Expanding this we get:

Perimeter = 8 x r x [cos (90 - (360/8))]

 

Let us now move away from the square to any perfect regular polygon like the one in the picture below.

http://img147.imageshack.us/img147/925/generalpolygx3.jpg

And now we generalize our experience from the square so that we can compute the perimeter of any regular perfect polygon with n sides using only r.

 

There are N pizza slices as shown in the picture above. The angle of the tusok of a single pizza slice is given by

360/N

Because theta is only half the value of the tusok of the pizza slice, the following gives the value of theta

theta = 360/2N

theta = 180/N

 

Again, we compute the value for phi by subtracting theta and 90 degrees from 180.

phi = 180 - 90 - 180/N

phi = 90 - 180/N

 

And we compute the value of a from r using trigonometry.

a = r x cos(phi)

a = r x [cos(90 - (180/N))]

 

In this general polygon, we see that the perimeter is composed of 2N a's. There are N pizza slices and there are two a's per edge of the pizza slice, therefore the total perimeter.

Perimeter = 2 x N x a

Perimeter = 2 x N x r x [cos(90 - (180/N))]

 

We now rearrange the variables

Perimeter = 2 x r x [N x cos(90-(180/N))]

This gives the PERIMETER OF ANY REGULAR POLYGON WITH N SIDES!

 

I now need you to look at the following picture. Each is a regular polygon. The white lettering refers to their number of sides.

http://img525.imageshack.us/img525/1285/polygonssn3.jpg

Do you notice that as the number of sides gets larger, the polygons start approaching the figure of perfect circles? We can then make this jump:

THE PERIMETER OF A CIRCLE IS EQUAL TO THE PERIMETER OF A REGULAR POLYGON WITH AN INFINITE NUMBER OF SIDES

But we already know how to derive the perimeter for any regular polygon!

Perimeter = 2 x r x [N x cos(90-(180/N))] (from above computations)

 

let us look at the bracketed term in that expression.

[N x cos(90-(180/N))]

I plotted the value of this expression from N = 1 to N = 1000 using MATLAB. The graph is shown below.

http://img137.imageshack.us/img137/938/n11000ng6.jpg

Notice that for very large values of N, the value of [N x cos(90-(180/N))] converges to a single value - 3.1415925. The value of

[N x cos(90-(180/N))] for very large values of N approaches pi!

 

We now rewrite our perimeter equation.

Perimeter = 2 x r x [N x cos(90-(180/N))] (Evaluated at very large values of N)

Perimerer = 2 x r x Pi

And thus we have derived the formula for the perimeter of a circle.

 

Whew! :thumbsupsmiley:

Edited October 23, 2007 by gineh

[bene_factor]

I don't think you can derive the area of a rectangle with sides L, W using integration because it is a circular argument.

 

When we take the "area under a curve" f(x) by integration, we are actually just adding the areas of little rectangles of height f(x) and width dx. So the starting point of the formulation already assumes the area of a rectangle.

 

But not to say the argument is not without value, but it is more of an exercise in demonstrating consistency.

 

--------------------------------------------

 

As for "pi", I believe the classical definition is that pi is the ratio of a circle's circumference to its diameter. Meaning, the formula C=pi.D=2.pi.r is a consequence of this definition.

 

 

 

All in all, I guess the two questions are related in that they ask about the origins of basic principles. In any mathematical system, one has to start with definitions and build a framework from there.

[Tank Girl]

The circumference of

a circle is 2 pi (about 6.28) times the radius of the circle.

Therefore the radius is the circumference divided by 6.28:

 

C = 2*pi*R

 

C/(2*pi) = R

[Tank Girl]

The circumference of

a circle is 2 pi (about 6.28) times the radius of the circle.

Therefore the radius is the circumference divided by 6.28:

 

C = 2*pi*R

 

C/(2*pi) = R

[Tank Girl]

The circumference of

a circle is 2 pi (about 6.28) times the radius of the circle.

Therefore the radius is the circumference divided by 6.28:

 

C = 2*pi*R

 

C/(2*pi) = R

[Tank Girl]

okay ive read the post earlier and how is it derived, i hope i can explain in simpler terms, math is such an intricate subject to discuss ( i love math!) first of all i have to discuss where it came from.. the CIRCLE, thats where it came from

 

Circle, in geometry, a two-dimensional curve such that each point on the curve is the same distance from a fixed point, called the center. The term circle may also be applied to the region enclosed by this curve. Simply said!

 

The circumference—distance around the edge—of a circle is equal to a constant, pi (symbol p), times the circle’s diameter: C = pd. Since the diameter of a circle is equal to twice the circle’s radius, the circumference also equals two times pi times the radius: C = 2pr. Pi is one of the most important mathematical constants, and plays a role in many calculations and proofs in mathematics, physics, engineering, and other sciences. The first ten digits of pi are 3.141592654, although the approximations 3.14 or 3 are sufficiently accurate for many calculations. 3. 14 and the common symbol, dont they teach these things in 6th grade?

 

Of all two-dimensional figures having the same perimeter, the circle has the greatest area. The area of a circle is equal to pi multiplied by the square of the circle’s radius: A = pr2.

 

To answer your question clearly i hope i answer it well is that this formula is derived from the Babylonians. Am i correct? Teehee! well someone has to give another math problem na!!!

[destron]

Great answers up there. ^^^

 

OK more of Math problems.

 

Sabong problem (again) :upside:

One day Quin went to 5 cockpit arenas. Each arena has an entrance fee of 5pesos and a winning bet gets 5 times the money bet. Everytime Quin entered an arena he plays only once, bets all his money, and wins. Also he gives a 5peso tip to the guard of the arena he just played. At the end of the day, he has 555 pesos.

 

How much money did he originally have?

Answer needs to be accurate in decimal places.

[ibirp]

orignal nomey = 7.6752

 

formula will be [(x-5) *5]-5 ---> reads as x (current money) minus 5 (entrace fee) then multiply the result to 5 (price you win) then subtract 5 from the result (tip to the guard).

 

so:

 

7.6752 - 5 = 2.6752; * 5 = 13.376; - 5 = 8.376 (first arena)

8,376 - 5 = 3.376; * 5 = 16.88; -5 = 11.88 (2nd arena)

11.88 - 5 = 6.88; * 5 = 34.4; -5 = 29.4 (3rd arena)

29.4 - 5 = 24.4; * 5 = 122; -5 = 117 (4th arena)

117 - 5 = 112 * 5 = 560; - 5 = 555 (5th arena)

 

this is actually done using manual steps (i worked backwards from 555 + 5 = 560; /5 = 112; + 5 = 117, and so on...), i'm still working for a formula for this...