Plzplzplzplz help me with these problems!

Heldensheld · 2006-07-23 01:20:42

I know, I know, I know, mods, this is like the nth time I posted for help with math....and no, n doesn't equal to blablabla infinity to the power of infinity !

Just that I am a bit slow and need help to find the answers. Trust me, i won't be sleeping when you people are working hard !

So please help me: !

Q1 Two circles of radii 4cm and 2cm touch a straight line at A and B as shown. Find the distance between A and B.

A_________________________B with circles on the line.

Q2 A rope is stretched tightly over an plane hangar from A to B. The shape of the hangar is semi-cylindrical. Find the length of the rope.
Shape is sorta like that with A at one end and B on the other.

(I have no clue on this one >.<, stupid weird roof!)

Q3 Solve the problem, "Two numbers have a sum of 1 and a sum of cubes of 31. Find their product!"

I gotta use the (x + y)^3 in it though I think.

Q4 Find a and b given that a^b x b^a = 800, where a and b are integers.

Thanks for your help peoples....One day...ONE DAY, I will help those who need help !

krassi_holmz · 2006-07-23 01:46:22

....One day...ONE DAY, I will help those who need help !

Promise???

Q3:
x+y=1
x^3+y^3=31
Now
(x+y)^3=1=x^3+3x^2y+3xy^2+y^3=
=(x^3+y^3)+3xy(x+y)=31+3xy.1, so
3xy+31=1
3xy=-30
xy=-10,
Now, by x+y=1, we have y=1-x:
x(1-x)=-10
x1=(1-√41)/2
y1=1-x1
x2=(1+√41)/2
y2=1-x2

Q4:

luca-deltodesco · 2006-07-23 03:09:00

Q1 doesnt really make any sense
Q2

circumference of circle = 2πr
so youre rope is stretched half way around, πr (π = pi = 3.141592653...bladibla)

but again, unsolvable, because you havnt told use what the radius of the cylinder is.

krassi_holmz · 2006-07-23 03:13:31

luca-deltodesco wrote:

Q1 doesnt really make any sense
Q2
circumference of circle = 2πr
so youre rope is stretched half way around, πr (π = pi = 3.141592653...bladibla)
but again, unsolvable, because you havnt told use what the radius of the cylinder is.

Abother person which thinks like me.

Heldensheld · 2006-07-23 14:24:47

Okay..Ill upload images.

[img]C:\Documents and Settings\Administrator\Desktop\5cerfib2[1].JPG[/img]

[img]C:\Documents and Settings\Administrator\Desktop\untitled.bmp[/img]

I can't load images from my own computer......

iiii
iiiiiii o
iiiiiiiiiii ooo
iiiiii ooooo
iiii ooo
______ii_____o_________________ Desparate time goes for desparate measures !
A B

Some one teach me how to upload?

Last edited by Heldensheld (2006-07-23 14:28:10)

Heldensheld · 2006-07-23 14:31:45

luca-deltodesco wrote:

Q1 doesnt really make any sense
Q2
circumference of circle = 2πr
so youre rope is stretched half way around, πr (π = pi = 3.141592653...bladibla)
but again, unsolvable, because you havnt told use what the radius of the cylinder is.

Wouldn't the radius be 25m because one of the lengths is 50m....and half that is 25m!

All_Is_Number · 2006-07-23 15:29:03

Heldensheld wrote:

Q1 Two circles of radii 4cm and 2cm touch a straight line at A and B as shown. Find the distance between A and B.
A_________________________B with circles on the line.

iiii
iiiiiii o
iiiiiiiiiii ooo
iiiiii ooooo
iiii ooo
______ii_____o_________________

Let us label the center point of circle A as point X. Let us label the center point of circle B as point Y. We will label the tangent point at which Circle A touches the line as point A. We will label the tangent point at which circle B touches the line as point B.

Draw a line segment from point X to point Y. We know this line segment has a length of 6 cm (4 cm + 2 cm = 6 cm).

Next, draw a line segment from point X to point A. Halfway between point X and point A on this line segment, create and label point Z.

Notice that line segment XZ has a length of 2 cm. Also notice that line segment YZ is parallel to the original line on which the circles lie, and is orthogonal to line segment XZ. Finally, notice that line segment YZ is equal in length to line segment AB.

We have created a right triangle XYZ with hypotenuse XY.

XY = 6 cm
XZ = 2 cm
YZ = n cm

By the Pythagorean theorem, 2^2 + n^2 = 6^2.

4 + n^2 = 36

n^2 = 32

n = sqrt(32) = 4 * sqrt(2)

~5.657 cm

I hope this helps.

Last edited by All_Is_Number (2006-07-23 16:09:52)

Heldensheld · 2006-07-23 18:32:12

Yayayaya !!!!!!!!

: jumps up n down like a crazy idiot :

It's right...im glad ppl undertand it -.-

Ricky · 2006-07-24 01:24:02

Draw a line segment from point X to point Y. We know this line segment has a length of 6 cm (4 cm + 2 cm = 6 cm).

Where does it say that the two circles touch at only one point? Or for that matter, even touch at all?

All_Is_Number · 2006-07-24 02:13:13

Ricky wrote:

Where does it say that the two circles touch at only one point? Or for that matter, even touch at all?

The "picture" seems to indicate the the circles share a single common tangent. If that is not the case, I do not believe there is enough information given to find a solution.

Heldensheld · 2006-07-24 02:30:59

I do have a picture but I dun know how to put it in without loading the picture of the internet.

All_Is_Number · 2006-07-24 02:47:50

Heldensheld wrote:

I do have a picture but I dun know how to put it in without loading the picture of the internet.

Do the circles touch each other at exactly one point? If not, my solution is incorrect and we need more information.

Heldensheld · 2006-07-25 02:53:23

Yes, they do....

Has anyone worked out Q2 yet ?

Last edited by Heldensheld (2006-07-25 02:53:37)

Patrick · 2006-07-25 02:57:44

Well, with the information we have atm, the answer is pi*r

All_Is_Number · 2006-07-25 04:33:47

Q2 A rope is stretched tightly over an plane hangar from A to B. The shape of the hangar is semi-cylindrical. Find the length of the rope.
Shape is sorta like that with A at one end and B on the other.

Since the rope is stretched tightly over the hangar, and point A is on one end of the hangar and point B is on the other end of the hangar, we can assume that point A and point B are on opposite corners of the hangar.

Again, the Pythagorean Theorem is the tool we need to solve the problem.

Let us call the length of the rope X.

Let us call the length of the hangar, from front to back L.

Let us call the width of the hangar at the base D.

D/2 = R

Note that R is the radius of the cylinder the hangar is based upon.

The Pythagorean theorem tells us: X^2 = (Pi *R)^2 + L^2

So X = sqrt((Pi * R)^2 + L^2)

Plug in the values for the variables (remember that R = D/2), and solve the expression.

Last edited by All_Is_Number (2006-07-25 04:35:30)

Heldensheld · 2006-07-26 02:51:12

R=25

L=Sqrt(50^2 +60^2)

All_Is_Number · 2006-07-26 06:27:55

Heldensheld wrote:

R=25
L=Sqrt(50^2 +60^2)

L is the length of the airplane hanger from front to back in my explanation. What is that length? Is it given as SQRT(50^2 + 60^2)?

X is the length of the rope.

BTW, what maths class is this for?

Last edited by All_Is_Number (2006-07-26 06:34:37)

Heldensheld · 2006-07-27 00:35:01

Highschool standards~

Math Is Fun Forum

#1 2006-07-23 01:20:42

Plzplzplzplz help me with these problems!

#2 2006-07-23 01:46:22

Re: Plzplzplzplz help me with these problems!

#3 2006-07-23 03:09:00

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#4 2006-07-23 03:13:31

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#5 2006-07-23 14:24:47

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#6 2006-07-23 14:31:45

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#7 2006-07-23 15:29:03

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#8 2006-07-23 18:32:12

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#9 2006-07-24 01:24:02

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#11 2006-07-24 02:30:59

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#14 2006-07-25 02:57:44

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#15 2006-07-25 04:33:47

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#16 2006-07-26 02:51:12

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#17 2006-07-26 06:27:55

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