Math Is Fun Forum

There's a slightly easier substitution that you can use for the second integral.  Let u = cos(5x) + 1.  Then you still have du = -5sin(5x)dx, giving you the integral

I'm not sure that I understand your question completely, but it sounds like you're looking for a way to match each natural number to a fraction, and vice versa.

If that is indeed what you are asking for, you're talking about the "countability" of different types of numbers.  As you pointed out, there are infinitely many natural numbers and infinitely many fractions.  However, we say that the set of fractions is "countably infinite" because we can create a function that maps every fraction to a natural number which is also invertible, meaning that we can create a bijection from the natural numbers to the fractions.  This Wikipedia article explains it further:

http://en.wikipedia.org/wiki/Countably_infinite

Sorry if this isn't what you were asking.

Answers 1, 2, and 4 can be eliminated quickly.  The mass of the sun is just less than 2*10^30 kg, which is 2*10^36 mg.  Even if we dramatically overestimate and say that the rest of the solar system has 9 times the mass of the sun we'd only end up with 2*10^37 mg of mass for number 4, which is quite small.  Numbers 1 and 2 will be much smaller yet than this number, eliminating them.

Number 7 is also quite a small number.  It's just 7 billion squared, which is about 4.9*10^19, again much too small for this question.

When we compare 5 and 6, it's simple to show that 6 is much larger than 5.  It's easy to show that 200! is much less than 200^200 (200! = 200*199*198*..., while 200^200 = 200*200*200*..., so each term is larger than it's factorial counterpart).  We know that 200^200 = 2^200 * 100^200.

Now, consider number 6.  The smallest such number will be 2 to a power that can be divided by 2, 3, 4, ..., 10.  The smallest such number can be found by factoring each number into it's primes and finding the smallest such number that includes every prime factor.  This number is 2^3 * 3^2 * 5^1 * 7^1 = 2520.  Thus, number 6 is 2^2520 = 2^200 * 2^2320.  All you must do now is convince yourself that 2^2320 is greater than 100^200.  Consider that 2^7 = 128 > 100, so we have 2^2320 > (2^7)^330 = 128^330 > 100^200.  Thus, we know that number 6 is a larger number than 1, 2, 4, 5, and 7.

That leaves only number 3.  As Ricky pointed out there is no time unit given.  The sun (and therefore Earth) take about 2.5*10^8 years to travel around the galaxy.  This is roughly 8*10^15 seconds.  Rounding down, the plank time is 10^-45 seconds.  That means that even if we use the smallest unit of time that has any physical significance it would only take a snail 8*10^60 time units to circle the galaxy, far less than 2^2520.  Thus, I declare number 6 the winner.

Yes.

http://en.wikipedia.org/wiki/Stirling%27s_approximation

Cleaning up the older questions:

1. We'll call the denominator d and the numberator n.  We know that the denominator is 3 less than twice the numerator, which mathematically means d = 2n-3.  We are also told that if 5 is added to both the numerator and denominator then the denominator becomes 1 more than the numerator, but this is the same as just saying that the denominator is 1 more than the numerator, which means d = 1+n.  Since we know the value of d from the second equation we can substitute its value into the first equation to get d = 2n-3  ==> 1+n = 2n-3.  You can now solve for both n and d.

2. We'll call the length l and the width w.  We're told that the length is 15 inches less than 3 times the width, whicih means l = 3w-15.  We also know that the perimeter of the painting is 98 inches.  Assuming that the painting is rectangular this means that twice the width plus twice the length equals the perimeter, or 2l+2w = 98.  Like before, we use substitution for l: 2(3w-15)+2w = 98  ==>  6w-30+2w = 98  ==>  8w = 128.  From here you can solve for both w and l.

3. Since we now know the length and width of the painting, multiply them to find the area in square inches.  Call the area A.  Your answer is simply 673,200 divided by A.

4. Multiply the ratios to find a common value.  Multiply the ratio of P to G by 6 to get 30:18, and multiply the ratio of P to M by 5 to get 30:20.  This means that for every 30 P books you have 18 G books and 20 M books.  Thus, the ratio of M to G is 20:18, which is equivalent to 10:9.

5. The average weight of 10 people is the sum of all their weights divided by 10.  Since their average weight is 145 pounds, their total combined weight is 1450 pounds.  When one person leaves that means that there are now 9 people with an average weight of 150 pounds, which means their total weight is 1350.  Subtract the weights and you'll find that the person that left weighed 100 pounds.

Again, no, for 2 of the reasons I mentioned in the other thread.  The cotton deforms, lengthening the amount of time that it's in contact with your body and therefore decreasing the amount of force it exterts on you at any given time.  Secondly, it has a larger surface area than the watermelon and spreads the force of the impact over a greater portion of your body, decreasing the amount of force exerted on your body at any given point.

1kg of watermelon is much more dense than 1kg of cotton, so when you get hit with the watermelon the force of the impact is much more concentrated than getting hit by the cotton.  That's why cutting tools are sharp: they allow the user to concentrate pressure along a very thin line or small point.

Also, as mathsyperson said, cotton is softer.  When it hits you it deforms, lengthening the amount of time that it's in contact with you.  Think about catching an egg or waterballoon.  You try to make initial contact with your hands away from you and then gently bring your hands into you, to maximize the amount of time that you can slow it down.  It's a matter of acceleration/deceleration.  The more time to have to accelerate/decelerate, the less force you need to apply at any given time.

Lastly, again due to the difference in density, the ball of cotton will be slowed down by air resistance much more than the watermelon.  It will be moving much more slowly by the time it reaches you.  Note that this particular property could be offset by doing the experiment in a vacuum, if you can just figure out a way to survive in such an environment.

I'm definitely rusty on double integrals, but I think you just evaluate one integral at a time and consider all other variables as constants.  For this problem, use substitution:

This substitution removes that ugly square root from the denominator, but we're left with an extra x.  Solving for x in terms of u gives us

We're now left to evaluate the following integral:

Evaluate the inside integral, pretending that y is a constant.  You should end up with a function in terms of y, which you then integrate over 0 to 1 for the final answer.

Edit: Or use Jane and Krizalid's method, it looks much more clean.