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(a^-9*b^-9)(a^3*b^-8)
= (1/a^9)(1/b^9)(a^3)(1/b^8)
= 1/(a^6*b^17)
1/(8xy^-6)^-1
= 1 / (y^6/8x)
= 8x/y^6
Well, Arthur C. Clarke sure thought it would work. He also thought we'd have intelligent AIs and the ability to send people to Saturn by 2001.
Suggestion for locking the hide tags:
Without altering the database in any way? Hmmm...It would be so easy if you could just add a toggle to the database...
If altering the database absolutely is not an option, there must be a place you could stick a toggle somewhere. In fact, you could stick it in a sort of "meta" tag in the post text. Give the post author a checkbox to turn it on and off, then put a hidden, never-output-to-the-browser bbcode tag along the lines of {enableHides=yes} in the message text and parse it like you do the rest of the bbcode tags.
edit: stupid typos.
Color me stupid, but I still don't understand how a counterweight is any different than a satellite in orbit, as far as the fact that it's not attached to anything, and therefore free to decelerate (and lose altitude and fall as a result) if a force is applied to it.
You may be right, mathsy.
I personally never understood how a satellite hanging in orbit is supposed to support the weight of a cable reaching to Earth, let alone a payload crawling up it. Wouldn't the downward force drag it out of orbit?
I have the Complete Idiot's Guide to Calculus, and it's a nice supplement to my calculus text.
It's just another substitution...
I figure they must be usefull if I'm being taught so much about it. We could teach kids in gradeschool to use calculators only and not how to do anything by hand. But first they wouldn't learn as much and they're mathskill would be entirely dependant on the calculator. They would be completely worthless without one.
I thought you might say that (everyone does). The answer is that both of our arguments are valid, and neither stands on its own. It's a question of where to draw the line.
Learning to integrate by hand is valuable as far as it goes, but there are some integrals that are just too tricky to be bothered with. This is why integral tables were invented, and later, calculators that integrate.
If I need to integrate for my job (Engineering, etc.), my employer will give me the tools I need to get the job quickly so I can focus on less mundane things. In this sense, an engineer is worthless without a calculator. Yet, engineering students are still taught how to do i.e. material balance problems by hand, even though there are advanced computer systems that do them for us. I think we both understand why that's valuable.
So, if an engineer is going to be using a computer to do tough integrals for him, does it make an ounce of difference whether he can do them by hand or not? Of course he should know how to do most integrals, but there are some...where do you draw the line?
Sorry to go OT on you. I'm better now. 8O)
I do, and I personally think it's more trouble than it's worth. But, it's not my forum.
You're homeschooled, right? Are you accountable to anyone for these problems? If so, ask them. If not, I say screw it, use a calculator.
Look, sane people use calculators. (I think I'll make that my sig.) Who wants to be puzzling over some bizarre integral when we've invented a tool to do it for us? Just working out the real-world problems that require integration is bothersome enough, and it's something no computer can do.
Oh yeah, and then we'd need a calculator to count to 10. Natural logs would be a breeze, though!
Google is your friend.
Also, mathforum.org has some good stuff, though it can be a bit frustrating to navigate.
Quantum computing might maybe have working prototypes in 20-30 years. Maybe.
Optical computing might also have potential as a silicon replacement, if they can succeed in trapping light on a small scale (yes, they've succeeded in trapping and storing a laser pulse for a time).
Back to the ionic drive probe: the advantage of ionic drive is that it produces very high exit velocities (around 70,000 mi/sec as opposed to chemical rocketry's 20,000 (IIRC)), so it's a very efficient conversion of electricity to propulsion. The trouble is twofold: one, the stream is very rarified, so that actual acceleration produced is slow. This means that we can't expect to power the thing with the sun, because the sun's rays will fade long before the probe is done accelerating.
The other problem is that even with the high exhaust velocities, energy density is still an issue. You have to carry enough propellent and energy (nuclear is currently the best candidate; most long-range probes have been nuclear powered) to reach those high velocities.
This is why the energy balance is more interesting to me than the time dilation; it's what's holding us back in space.
The "post Ender's Game" trilogy was pretty much based on this idea.
To me, the more interesting problem is the energy balance. How much energy does it take to accelerate to 99% of c and back down again? I have read that the only known energy source with enough energy density to pull it off is antimatter.
Hmm...I think that would require relativistic physics to figure out, and as a ChemE I don't think I'll ever get there.
Showoff.
"What do you mean, I can only weigh one time? Forget that. I'll find a job at a factory that isn't managed by lunatics."
Anyway, I think Rod's got it. If you select as your sample sizes successive powers of two, then weigh the lot. Subtract the weight from what it would be if all were running smoothly (10230), convert the result to binary, and you'll be able to to tell based on which bits are set which machines are working properly or not.
1) It's (8x - 3)(8x + 3).
2) Let n be the number you're looking for. Then:
n² + (n+1)² = 61
n² + n² + 2n + 1 = 61
2n² + 2n - 60 = 0
2(n-5)(n+6) = 0
n = -6 and n = 5.
So, both -6, -5 and 5, 6 will work.
3) (3z+2)(5z-4)
4) (5x-3)² = 18x² + 1
25x² - 30x + 9 = 18x² + 1
7x - 30x + 8 = 0
(x-4)(7x - 2) = 0
x=4 and x = (2/7)
The integer solution is x=4.
5) The total area of the page is 12*18 = 216 cm². The printed material covers 40 cm², leaving 176 for the margins. Set up an equation for the margins and solve it.
Let w = the width of the margins. Then,
12w + 12w + 2(18-2w)w = 176
-4w² + 60w - 176 = 0
-4(w-11)(w-4) = 0
w = 11 or w = 4
Two widths would work, but 11 is almost 12, which is one of the dimensions of the page, so that's probably not the best choice. Let's check it with 4:
48 + 48 + 40 + 40 = 176
6) You're correct on this one.
Note that I used a calculator to factor some of those quadratics, since factoring isn't one of my strong points. Lacking that, I'd resort to the quadratic formula.
Note also that when factoring, I always have to expand my factored expression to check and see if it's right.
I loved the ski jumper. It was the most amazing bit of coordinated choreography I've ever seen.
I always quit when the countries start coming out, though. So. Boring. Did they do anything after that?
If you're just talking about bases, you can base a number system on any arbitrary base you want. It just depends on what's convenient for you. Computers use base-2 not because it's awesome, but because they don't have much of a choice considering they only have two states. (Whoever thought up binary and applied it to computers was a genius...)
The Babylonians used a base-60 number system. Yikes...how did they get anywhere with base 60?
Find someone in your class that has the CD and copy it.
Who said anyone had a fish?
Some people are asking for a challenge...
Or maybe you're just jealous of his different kind of cool?