Math Is Fun Forum

Who knows, man. Homework with no instructions is one of my greatest pet peeves with school.

My best guess is they want you to categorize the shapes under headings such as triangles, squares, rectangles, polygons, conics, etc.

The plus is that, with instructions that vague, the number of answers that work is much greater. There's no one "right answer" for each group, because if every shape in the group fits the bill, then your answer is correct.

The equation y = mx + b will serve you well here. This is called "slope-intercept' form of the equation for a line, because m represents the slope and b represents the point where the line crosses the y axis, or y-intercept.

I think it's safe for us to assume that the origin O corresponds to the point (0, 0) on the coordinate plane. So, a line that passes through the origin will have a y-intercept of 0; that means b=0 and we don't have to worry about it.

Now, we need to find the slope. This is "rise over run". So, A (1, 2) is 2 above O and 1 to the right, so m = 2 / 1 = 2.

The final equation is y = 2x. You can verify this by plugging in a number (1) for which you know the answer (2), like so: y = 2(1) = 2. Bingo!

Now, a perpendicular line has a slope that is the negative reciprocal (flip the fraction over and negate it) of the first line. So, the slope of our new line is (-1/2). Now our equation is y = (-1/2)x + b. We need to find the value of b, so we need numbers to plug into our other variables. We have them, since the line (AB) passes through our given point (1, 2). So, solve the equation 2 = (-1/2)(1) + b for b, and you can find b and your final equation.

The line will cross the x-axis where y = 0, so set y equal to 0 in your equation and solve for x to find the coordinates.

Hey sony, what happens if you have a sphere? There's technically no facets...

You took 2 minutes of my life and I WANT THEM BACK!

Seriously, though, the Mayas had a calendar that didn't need all this leap garbage. We should toss out the baggage of centuries, ditch the gregorian/julian calendar, and go to a sensible system based more on the actual motions of the earth than our desire to have a neat, integral number of hours in a day (or days in a year).

Like that'll ever happen...

The cosine and sine functions are periodic, so they have infinitely many maximums. The period of both is pi radians (180 degrees), though cos begins pi/2 radians to the left of sin. So, at every increment of pi/2, either sin or cos has a maximum.

You can use this info to figure out the answer...

"^" means to raise to a power. It's necessary on plain text documents that don't support superscripts. It's also an operator in many programming languages, including, I think, C/++, so you can use it in your programs to raise things to a power.

Thanks.

Me.

Inverting words is a hobby of mine. I can't claim to be tremendously good at it, but I'm gratified that you could read it and recognized its significance.

The man whose work got me started is Scott Kim. His are still the most brilliant I've seen. His book, "Inversions," is also a fascinating read; it has a ton of his drawings in it, too.

Wow. It's been a really long time since I did any C programming, and I don't remember what Math functions are available. So, I'll give a couple simple examples in Java, which has similar syntax.

The Quadratic Formula:

public Point quadform (a, b, c)  {
       float plus, minus;

       plus = (b^2 + Math.sqrt (2*b - 4*a*c)) / (2*a);
       minus = (b^2 - Math.sqrt (2*b - 4*a*c)) / (2*a);

       return new Point (plus, minus);
}

n!:

public double factorial (n)  {
       double answer = 2;

       if (n > 1)  { answer = factorial (n-1); }
       
       return answer * (n-1);
}

This may sound really stupidly obvious, but writing algorithms of math functions is just programming. Get a book and practice. If you're having a problem with a specific function, post a specific question and we'll be happy to answer.

Thanks for the replies. Now I get where the 2 comes from. Now my question is: where does the du go? u²/2 is the integral of u...what happened to the du?

Thanks again.

We are constrained by physical laws which make drilling the hole impossible unless it turns out we were wrong about those laws. You are limited by materials and energy. It's like trying to build a fusion reactor. We have been able to produce fusion for 50 years or so, but only in bombs--we can't contain it. The properties of matter say it's impossible to produce a material that can withstand the temperatures of fusion without ionizing and turning into plasma themselves. So you need to generate a magnetic field strong enough to keep it from touching anything, and we haven't been able to do that yet.

Or space travel. We pretty much reached our limits on space travel technology in the 60's and 70's, because there does not exist a fuel with sufficient energy density to propel a passenger ship at a reasonable rate. It takes our unmanned probes years to reach Jupiter. The only candidate is antimatter, which we cannot produce or store with near the efficiency required for practical application.

Talking about the hole, there doesn't exist a material with sufficient tensile strength to extend from the surface of the earth to the center--it would break under its own weight. That is, unless we discover one (buckminsterfullerine looks promising).

If we could make an antigravity field to balance the weight, there's still the problem of having a motor big enough to turn it. Even with a nuclear-powered gas turbine, I'm not sure you could produce enough power.

And if you had an antigravity field, how much power would it require? Could you project it along the length of the tunnel?

All of this is ignoring the presence of magma.

A thought just occurred to me. We could perform this experiment on a smaller scale in space. Not that we'd want to...

Maths - I may be reading your code wrong, but I don't think that's quite the effect he was looking for.

In order to constrain the proportion to 4:3, you need to change one side in proportion to the side whose delta is greater. That way, the user's mouse stays connected with at least one side of the rectangle. So:

if (deltaX > deltaY)  {
       deltaX = (3/4) * deltaX
}
else  {
       deltaX = (4/3) * deltaY
}

However, that algorithm falls down when the user drags the mose above or to the left of the top left corner of the rectangle, since the deltas become negative. So, we need to work in terms of the coordinates of the corners, instead of the width/height of the rectangle.

Declare variables  (xTL, yTL);  (xBR, yBR);  to hold the coordinates of the top left and bottom right corners of the rectangle. xM and yM refer to the position of the mouse. The deltas will also make an appearance.

if (xM > xBR)  {
     deltaX = xM - xBR;
     deltaY = yM - yBR;
     
     if (deltaX > deltaY)  { xBR = xM;  yBR = (3/4) * xM; }
     else  { xBR = (4/3) * yM;  yBR = yM; }

     if (yM < yTL)  {  //The user is above the top left; invert!
          yBR -= 2 * (yBR - yTL);
     }
}

if ((xM < xBR) && (yM > yBR))  {
     //The user is below and to the left of the bottom right
     //Adjust yBR and xBR accordingly, and
     //make sure to check for the user being to the left of xTL
}

And keep going until you've covered every possible situation. Sorry I ran out of time and couldn't finish it.