forward
Smash Champion
16^x = 1/sqrt2
VVVVVV
4^2x = 1/sqrt2
How do I make the bases equal? I believe that's the next step.
VVVVVV
4^2x = 1/sqrt2
How do I make the bases equal? I believe that's the next step.
-
Welcome to Smashboards, the world's largest Super Smash Brothers community! Over 250,000 Smash Bros. fans from around the world have come to discuss these great games in over 19 million posts!
You are currently viewing our boards as a visitor. Click here to sign up right now and start on your path in the Smash community!
Need Math Help?
- Thread starter AltF4
- Start date
Are you just trying to solve for x? If that's the case, then just take the log of both sides.
And what does your "VVVVV" mean? Is that meant to imply that you started with "16^x = 1/sqrt2" and then changed it to "4^2x = 1/sqrt2"? If that's the case, then there probably isn't any need to do that.
You could just take the log base 16 of both sides to "16^x = 1/sqrt2". Your answer would be
X = log16(1/sqrt2)
1UPChris:
The wikipedia page is pretty good, I think. Standard form is just another way of writing the equation to the line. Unfortunately, it's not as useful as others in most cases. Point-Slope form and Slop-Intercept form you tend to see more often.
I'm not sure what there really is else to understand about it.
Perhaps you could be more specific?Smash Noob
If you want integration answers, go here. It's pretty good. But of course you still have to figure it out on your own, too!
As for your equation, it seems to integrate out to something really wierd. Are you sure that's the right equation?
forward
Smash Champion
So i'm using all these formulas with e. Why does e represent constant change? Like compound interest formulas, Newton's law of cooling, etc.
Excellent question. But one that is not simple to answer. The number e is a "transcendental" number. Which (loosely) means that the value of that number is built into the laws of mathematics itself. The way that exponential growth and decay work naturally creates that number. It's really fascinating!
Another famous transcendental number is Pi. No matter what universe you're in, the ratio of a circle's circumference to it's diameter will be be Pi.
There's no reason "why". There's no way to say "this is how we came up with that number". That number is just intrinsically built into the universe itself.
'e' happens to be special when talking about exponents for reasons that won't be obvious until calculus. It happens to be the only base for an exponent which is itself when integrated. It is the "identity" case for exponentiation integration. (Much in the same way that anything times 1 is itself, anything with a base of e will be itself when integrated)
Another famous transcendental number is Pi. No matter what universe you're in, the ratio of a circle's circumference to it's diameter will be be Pi.
There's no reason "why". There's no way to say "this is how we came up with that number". That number is just intrinsically built into the universe itself.
'e' happens to be special when talking about exponents for reasons that won't be obvious until calculus. It happens to be the only base for an exponent which is itself when integrated. It is the "identity" case for exponentiation integration. (Much in the same way that anything times 1 is itself, anything with a base of e will be itself when integrated)
Are you just trying to solve for x? If that's the case, then just take the log of both sides.
And what does your "VVVVV" mean? Is that meant to imply that you started with "16^x = 1/sqrt2" and then changed it to "4^2x = 1/sqrt2"? If that's the case, then there probably isn't any need to do that.
You could just take the log base 16 of both sides to "16^x = 1/sqrt2". Your answer would be
X = log16(1/sqrt2)
1UPChris:
The wikipedia page is pretty good, I think. Standard form is just another way of writing the equation to the line. Unfortunately, it's not as useful as others in most cases. Point-Slope form and Slop-Intercept form you tend to see more often.
I'm not sure what there really is else to understand about it.
Smash Noob
If you want integration answers, go here. It's pretty good. But of course you still have to figure it out on your own, too!
As for your equation, it seems to integrate out to something really wierd. Are you sure that's the right equation?
Thanks, I checked out wikipedia, and it helped me finish my project.
1048576
Smash Master
- Joined
- Oct 1, 2006
- Messages
- 3,417
So, I was mulling over a problme in my head, but I gave up, so I'm wondering if you guys can help. It's not homework or anything, I'm just genuinely curious.
The way the NFL overtime system works, the referee flips a fair coin. Whoever wins the coin toss gets to have the first chance to score. If they fail to score, the other team gets a chance, then if they fail to score, the first team gets another chance, and so on. Whoever scores first wins. Suppose you and I are equally skilled, and we end up in overtime under NFL rules. You win the coin toss. If both of us have an X probability of scoring on any given chance, what is the probability (in terms of X), that you win?
The way the NFL overtime system works, the referee flips a fair coin. Whoever wins the coin toss gets to have the first chance to score. If they fail to score, the other team gets a chance, then if they fail to score, the first team gets another chance, and so on. Whoever scores first wins. Suppose you and I are equally skilled, and we end up in overtime under NFL rules. You win the coin toss. If both of us have an X probability of scoring on any given chance, what is the probability (in terms of X), that you win?
iirc, there isn't actually any significant correlation between the coin toss and games won. But this is complicated by the fact that in football your defense can score and win the game. Indeed depending on the matchup it might be in your best interest to let the other team take the ball!
But ignoring that and just talking about odds, it should be pretty straight-forward. Let's just assume that there's a 50% chance of a team scoring on its drive.
Team 1 starts with a 50% chance to win right off the bat. So half of all overtime games would end with the receiving team scoring on the opening drive.
If not, Team 2 then has a 50% chance of scoring. So one quarter of all games would end with Team 2 scoring.
One eighth of games would end with Team 1 getting the ball back and scoring. etc...
so the formula looks like SUM [ (1/2)^n ] (for odd n's) for Team 1
and SUM [ (1/2)^n ] (for even n's) for Team 2.
I don't have a calculator in front of me, and I have class in a moment. But I'm sure you could try to find the actual odds. (It appears to be around 70% total win rate for the team that starts first)
But ignoring that and just talking about odds, it should be pretty straight-forward. Let's just assume that there's a 50% chance of a team scoring on its drive.
Team 1 starts with a 50% chance to win right off the bat. So half of all overtime games would end with the receiving team scoring on the opening drive.
If not, Team 2 then has a 50% chance of scoring. So one quarter of all games would end with Team 2 scoring.
One eighth of games would end with Team 1 getting the ball back and scoring. etc...
so the formula looks like SUM [ (1/2)^n ] (for odd n's) for Team 1
and SUM [ (1/2)^n ] (for even n's) for Team 2.
I don't have a calculator in front of me, and I have class in a moment. But I'm sure you could try to find the actual odds. (It appears to be around 70% total win rate for the team that starts first)
1048576
Smash Master
- Joined
- Oct 1, 2006
- Messages
- 3,417
I know that it's 2/3 for a 50% chance, and that it's 4/5 for a 75% chance, but I'm having issues for the 25% chance and the general rule.
I don't think that formula works for any value other than 1/2, because for other values, the "remaining probability up for grabs" is not the value itself raised to some power.
I don't think that formula works for any value other than 1/2, because for other values, the "remaining probability up for grabs" is not the value itself raised to some power.
Proverbs
Smash Lord
I hate your signature. SO MUCH.
yea im glad I found this- im having a slight issue understanding Quaternions.
I understand the basic purpose and need, I just am stuck on a roadblock with the simple math of it.
anyone feel like throwing some ideas my way, feel free.
I understand the basic purpose and need, I just am stuck on a roadblock with the simple math of it.
anyone feel like throwing some ideas my way, feel free.
INSANE CARZY GUY
Banned via Warnings
just make yourself a basic system. like when i need to * something by 9 i take the # lower than the number i am working with like 7*9 =6 and then i take 10-7 to get 3 "63" just make system like that
now my problem http://www.smashboards.com/showthread.php?p=6961768#post6961768
now my problem http://www.smashboards.com/showthread.php?p=6961768#post6961768
i am praying that wasnt directed at me...... because that gives me nothing...
c3gill:
Oh, my. I'm afraid I think that's a bit beyond me. I only got a minor in mathematics, after all. It's not my full-time job. Perhaps if you were a bit more specific in what you need help with?
As always, Wikipedia is the first place you should look. So if you haven't already...
http://en.wikipedia.org/wiki/Quaternion
Oh, my. I'm afraid I think that's a bit beyond me. I only got a minor in mathematics, after all. It's not my full-time job.
Perhaps if you were a bit more specific in what you need help with?As always, Wikipedia is the first place you should look. So if you haven't already...
http://en.wikipedia.org/wiki/Quaternion
lol yea, im having an issue with some of the basic algebraic functions and how to properly represent them in complex planes.c3gill:
Oh, my. I'm afraid I think that's a bit beyond me. I only got a minor in mathematics, after all. It's not my full-time job.
As always, Wikipedia is the first place you should look. So if you haven't already...
http://en.wikipedia.org/wiki/Quaternion
Basically, if you have a decent understanding of Quaternion Algebra, please PM me! I could use some help!
metalmonstar
Smash Lord
- Joined
- Apr 30, 2008
- Messages
- 1,081
When doing double integrals, how do you know when to integrate with respect to y/x first?
Corpsecreate
Smash Lord
I spent a few hours trying to do this one last night but I eventually got it.
Integrate Sin(x)Cos(x) / (1-Cosx)
The final answer I got was Ln(Sinx) + Ln[ (1 / Sinx) + (1 / Tanx)] + Cosx
I know that this answer is correct but think of it as a challenge to anyone that might wanna try it
Integrate Sin(x)Cos(x) / (1-Cosx)
The final answer I got was Ln(Sinx) + Ln[ (1 / Sinx) + (1 / Tanx)] + Cosx
I know that this answer is correct but think of it as a challenge to anyone that might wanna try it
what technique did you use, i tried let u = 1-cosx, but i got a diffrent answer ln(1-cosx) +1 -cosxI spent a few hours trying to do this one last night but I eventually got it.
Integrate Sin(x)Cos(x) / (1-Cosx)
The final answer I got was Ln(Sinx) + Ln[ (1 / Sinx) + (1 / Tanx)] + Cosx
I know that this answer is correct but think of it as a challenge to anyone that might wanna try it
Corpsecreate
Smash Lord
I simplified the expression and applied some trig identities before trying to integrate it.
KosukeKGA
Smash Champion
Bob takes his girlfriend to the fair's Merry-Go-Round. If the axle turns at 6 revolutions/minute, and the rider is 12 ft from the axle, how fast is the rider's velocity?!?
I think you must mean "speed", not "velocity".
In order to determine speed, we need to know two pieces of information:
1) How much distance was traveled
2) How much time it took to travel that distance
To find #1, we use a bit of geometry. The merry-go-round is presumably a circle. So if the merry-go-round has a radius of 12 feet, then it will have a circumference of 2* Pi * 12 = 24 Pi
(Circumference = 2 * Pi * radius)
Therefore, every revolution the merry-go-round makes, the person will travel 24 Pi feet. The problem supposes that she travels 6 revolutions, however. So the full distance traveled is 144 Pi.
All of this was done in one minute. But let's call it 60 seconds so that we get our answer in ft/sec.
We then just need to divide the distance by the time and we get:
144 Pi / 60 feet per second, or around 7.539 feet per second.
In order to determine speed, we need to know two pieces of information:
1) How much distance was traveled
2) How much time it took to travel that distance
To find #1, we use a bit of geometry. The merry-go-round is presumably a circle. So if the merry-go-round has a radius of 12 feet, then it will have a circumference of 2* Pi * 12 = 24 Pi
(Circumference = 2 * Pi * radius)
Therefore, every revolution the merry-go-round makes, the person will travel 24 Pi feet. The problem supposes that she travels 6 revolutions, however. So the full distance traveled is 144 Pi.
All of this was done in one minute. But let's call it 60 seconds so that we get our answer in ft/sec.
We then just need to divide the distance by the time and we get:
144 Pi / 60 feet per second, or around 7.539 feet per second.
KosukeKGA
Smash Champion
Ho, hum.
Thank you, Alty.
---
Blame my math teacher for mixing the two terms up. -.-
Thank you, Alty.
---
Blame my math teacher for mixing the two terms up. -.-
Well, they're very related.
Speed is a scalar value. Meaning it's just a number. So you can have a "speed" of 50 miles per hour.
But Velocity is a vector. It is the combination of both speed AND direction. In order to find your velocity, you also have to find the direction you're moving.
So in your merry-go-round case, the person's velocity is constantly changing! The person's speed remains the same, sure. But the direction he's moving keeps changing.
And what do we call a change in velocity? Acceleration. That's why it feels like the merry-go-round is trying to throw you off. Because you're accelerating. It's the same as if you put your foot down on the gas pedal in a car.
Speed is a scalar value. Meaning it's just a number. So you can have a "speed" of 50 miles per hour.
But Velocity is a vector. It is the combination of both speed AND direction. In order to find your velocity, you also have to find the direction you're moving.
So in your merry-go-round case, the person's velocity is constantly changing! The person's speed remains the same, sure. But the direction he's moving keeps changing.
And what do we call a change in velocity? Acceleration. That's why it feels like the merry-go-round is trying to throw you off. Because you're accelerating. It's the same as if you put your foot down on the gas pedal in a car.
Frown
poekmon
756=(6x-9)*x
Moon of the Strawberries
Smash Journeyman
Hey, I need a favor from someone. I'm going to be taking a math quiz on Monday and we're doing roots, rational exponents, powers, roots and radicals etc etc(i'm in Algebra 2). Can anyone give me a few hints/tips on doing exquations like these?
1048576
Smash Master
- Joined
- Oct 1, 2006
- Messages
- 3,417
nvm, I solved my problem.
Um, that seems too vague to be useful. Remember to not leave variables in the denominator. If you end up with a variable in the denominator, which happens with negative exponents, multiply both sides by that expression.
Um, that seems too vague to be useful. Remember to not leave variables in the denominator. If you end up with a variable in the denominator, which happens with negative exponents, multiply both sides by that expression.
When you say "variables in the denominator" I think you mean unresolved radials (aka surds), not variables. And in any case, there's no good reason not to leave radials in the denominator.
One mistake that beginners occasionally make is assuming that exponential is linear; it isn't.
(a + b)^x != a^x + b^x.
I'd watch out for that if you are new to powers.
As for solving equations, make sure you take note of the fact that squaring and taking the principal square root are not inverse operations; you need to consider the sign of the expression under the radial.
One mistake that beginners occasionally make is assuming that exponential is linear; it isn't.
(a + b)^x != a^x + b^x.
I'd watch out for that if you are new to powers.
As for solving equations, make sure you take note of the fact that squaring and taking the principal square root are not inverse operations; you need to consider the sign of the expression under the radial.
illinialex24
Smash Hero
Yeah (a+b)^x is actually a fairly tricky expansion.
It follows a binomial distribution and it goes like this
XC0 a^X + XC1 a^x-1*b^1...... XCx-1 a^1*b^x-1 + XCX b^x
Thats your expansion pattern
It follows a binomial distribution and it goes like this
XC0 a^X + XC1 a^x-1*b^1...... XCx-1 a^1*b^x-1 + XCX b^x
Thats your expansion pattern
1048576
Smash Master
- Joined
- Oct 1, 2006
- Messages
- 3,417
No actually I meant if you get something like 1/x^3 = 27, multiply both sides by x^3. When I was in middle school, I'd encounter something like that, get stuck, and start doing guess and check. Now I know how to properly deal with it.
Moon of the Strawberries
Smash Journeyman
Ok, I have a math question i'm having a bit of trouble with.
It's 4^1/4 x 64^1/4 and I have to simplify it. For some reason i'm having a hard time with it.
It's 4^1/4 x 64^1/4 and I have to simplify it. For some reason i'm having a hard time with it.
Wow, I'm completely lost...
Three circles with centres A, B, and C are mutually tangent and no circle lies inside another circle. The circle with centre A has radius 3 cm and the circle with centre B has radius 5 cm.
a) If angle BAC = 60 degrees, then determine the radius of the third circle.
b) Find the area of triangle ABC.
Okay so I got a = 7 cm and thats about it?
Three circles with centres A, B, and C are mutually tangent and no circle lies inside another circle. The circle with centre A has radius 3 cm and the circle with centre B has radius 5 cm.
a) If angle BAC = 60 degrees, then determine the radius of the third circle.
b) Find the area of triangle ABC.
Okay so I got a = 7 cm and thats about it?
Nitrix
Smash Ace
Could somebody help me solve this optimization problem?
A metal frame in the shape of a right triangle has sides of length 6cm, 8cm, and 10cm. A rectangular plate will be welded to the frame, with two eges of the plate attached to the sides of the frame, and one corner of the plate welded to the hypotenuse. A contractor needs an estimate of the amount of material needed for the plate. What is the maximum area of the rectangular plate?
The triangle has a 10cm hypotenuse, the base is 8cm, and the height is 6cm.
A metal frame in the shape of a right triangle has sides of length 6cm, 8cm, and 10cm. A rectangular plate will be welded to the frame, with two eges of the plate attached to the sides of the frame, and one corner of the plate welded to the hypotenuse. A contractor needs an estimate of the amount of material needed for the plate. What is the maximum area of the rectangular plate?
The triangle has a 10cm hypotenuse, the base is 8cm, and the height is 6cm.
Corpsecreate
Smash Lord
.------------------- .______________
.-------------------/|------------------|
.------------------/-|-----Area x------|---- <--- Plate
.-----10cm-----/--|------------------|
.----------------/---|-6cm------------|
.---------------/----|------------------|
.--------------/____|____________ |
.---------------8cm
Am I seeing this right? Btw drawing this was hell
.-------------------/|------------------|
.------------------/-|-----Area x------|---- <--- Plate
.-----10cm-----/--|------------------|
.----------------/---|-6cm------------|
.---------------/----|------------------|
.--------------/____|____________ |
.---------------8cm
Am I seeing this right? Btw drawing this was hell
Nitrix
Smash Ace
The plate is actually inside the triangle, with one side resting on the 6cm side, and the other one resting on the 8cm side. The corner of the rectangle is touching the hypotnuse.Am I seeing this right? Btw drawing this was hell
Thanks for your help!
Corpsecreate
Smash Lord
Ah I understand whats it supposed to be now. Its easy
Ok. first thing you need to do is get the dimensions of the rectangular plate inside the triangle in terms of x.
Let x be the part of the triangle that you are "cutting out" from the bottom side (6cm). Now you should know that if you half the base of the triangle then the height will also be halved. So if you take off x from the base, the height will become an x/6th of the original height. So now you have both the Length and the Width of the plate in terms of x.
Length = 8(x/6)
Width = (6-x)
Area = (6-x) . 8(x/6) = (48x/6) - (8x²/6)
Differentiate this and you get: 8 - 16x/6
8 - 16x/6 = 0 for maximum therefore x = 3.
Largest possible area is then (6 - 3) x 8(3/6) = 12cm²
Hope that helps!
Now I have a question I need help with
Lim
x-->∞ (X/X+1)^X
Question says I have to use L'Hopitals rule to solve it. I'm pretty sure the answer is either 1 or 0, I can understand why it would be either of them but im not sure which one is correct.
I think its 1 because ∞+1 = ∞.
so... (∞/∞+1)^∞ = (∞/∞)^∞ = 1^∞ = 1
But then, it makes sense for it to be zero because ∞+1 > ∞ (well it sort of is), so (∞/∞+1)^∞ = 0^∞ = 0
Of course this isnt using L'Hopitals rule but by using it, you end up with a similar situation.
Ok. first thing you need to do is get the dimensions of the rectangular plate inside the triangle in terms of x.
Let x be the part of the triangle that you are "cutting out" from the bottom side (6cm). Now you should know that if you half the base of the triangle then the height will also be halved. So if you take off x from the base, the height will become an x/6th of the original height. So now you have both the Length and the Width of the plate in terms of x.
Length = 8(x/6)
Width = (6-x)
Area = (6-x) . 8(x/6) = (48x/6) - (8x²/6)
Differentiate this and you get: 8 - 16x/6
8 - 16x/6 = 0 for maximum therefore x = 3.
Largest possible area is then (6 - 3) x 8(3/6) = 12cm²
Hope that helps!
Now I have a question I need help with
Lim
x-->∞ (X/X+1)^X
Question says I have to use L'Hopitals rule to solve it. I'm pretty sure the answer is either 1 or 0, I can understand why it would be either of them but im not sure which one is correct.
I think its 1 because ∞+1 = ∞.
so... (∞/∞+1)^∞ = (∞/∞)^∞ = 1^∞ = 1
But then, it makes sense for it to be zero because ∞+1 > ∞ (well it sort of is), so (∞/∞+1)^∞ = 0^∞ = 0
Of course this isnt using L'Hopitals rule but by using it, you end up with a similar situation.