Matrices FTL!Determinants are something entirely different.
Matrices FTL!Determinants are something entirely different.
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What's the difference between microfilaments and microtubules? (In cellular biology)
Well, off the top of my head, microfilaments are made out of actin, while microtubules are made from tubulin. In terms of actions, microfilaments make up the bulk of the cell's cytoskeleton, and the depolarization and repolarization of them is what allows cells like macrophages to move to where they are needed. Microtubules, on the other hand, make up structures such as the mitotic spindle, as well as things like flagella (they tend to be arranged in a 9x2 circular arrangement).
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so microtubules aren't considered part of the cytoskeleton?
Edit: sorry, a search in wikipedia shows that microtubules and microfilaments are both considered part of the cytoskeleton.
Last edited by Gungalley; August 24, 2007 at 04:52 AM.
This one's aimed at the people who know some basic maths.
f(x) = 2x-1, g(x) = 3-2x and h(x) = ¼(5-x)
a) Find a formula for k(x) where k(x) = f(g(x))
b) Find a formula for h(k(x))
c) What is the connection between the functions h and k?
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a) k(x)=2(3-2x)-1 =5-4x
b) h(k(x))=(1/4)(5-(5-4x)) =(1/4)(4x) =x
c) hmm, not sure... been a while
better figure it out soon though... gonna tutor someone today in this type-o math
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x is just a symbol, that can be anything. Let's use two different symbols for clarity. We know that for any x and any y, we have
g(x) = 3 − 2x
f(y) = 2y − 1
Since y can be anything, let's choose y = g(x). Then we just have to plug it into the formula:
k(x) = f(g(x)) = 2g(x) − 1 = 2(3 − 2x) − 1 = 5 − 4x
All you have to realize is that x can be anything in that formula, including an expression that contains x itself! Then you just have to substitute, the essence of mathematics.
Same thing:
k(x) = 5 − 4x
h(k(x)) = (5 − k(x))/4 = (5 − (5 − 4x))/4 = x
This one requires more thought, since it's not strictly a mathematical question. You have two functions h and k, so that h(k(x)) is equal to x itself. Now, if I told you I had two numbers a and b so that a × b = 1, what's the relationship between them? If instead a + b = 0, what's the relationship?
The answer is the same in all cases: that of inverse. If a + b = 0, then b = −a, its additive inverse. If a × b = 1, b = 1/a, its multiplicative inverse. And if a(b(x)) = x, then b(x) = a^{−1}(x), its compositional inverse. In each case, performing an operation between two values gives you back the identity, a value that doesn't change what it operates on. If f(x) = x, then g(f(x)) = f(g(x)) = g(x) for any g. Likewise, if a = 1, then a × b = b for any b, and if a = 0, a + b = b for any b. If you're being asked questions like this, you probably want to remember what the identity for composition is, too.
Thanks Sim and Drak!
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Just one more question (for now ):
2.5 = 2+1sinx (this is radiens, not degrees), I need to find all the values in pi where y = 2.5, I suppose it's similar to y = 2sinx - it is just that slid up 2 units in the y-axis. Thanks in advance.
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ahh, just erased everything I wrote! so here goes again:
well, I guess your question is:
for a function y=2+sin(x) find all the values of x for which y=2.5, right?
Now, first, we can simplify this, because all you have to do is solve for sin(x)=1/2
because:
2.5=2+sin(x)
=> 2.5-2=sin(x)
=> .5=sin(x)
Now, you just have to remember (or look) on your unit circle, for when sin(x)=1/2. If you don't remember it, here it is:
(remember, that you just have to know the first quarter of the unit circle to find the rest of the values on it)
so, remember that sin(x) give the "y" value for a particular rad/deg. So if you look at the circle, sin(x)=1/2 when x=(pi)/6... also, it is equal to 1/2 when x=(5pi)/6. You must also remember, that if you add 2pi to either of those values, you will also obtain the same answer, since 2pi is 360deg and you are just going around in a circle coming back to the same location. So technically, there are an infinite amount of answers...
Your final answer should be something like this, though I'm not sure what notation you guys use, but in cal I think we usually wrote it as:
sin(x)=1/2 when x=(pi)/6 + 2(pi)i and x=(5pi)/6 + 2(pi)i (where "i" is any positive or negative integer.
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hmm, having a bit of trouble with this one... and the teacher did not provide an answer. Its not really that important, and the test is tomorrow, but I'm just interested.
A,(BxC)=-4 (where "," is the dot product, and "x" is the cross product)
find:
A,(CxB)
(BxC),A (this one is obvious)
C,(AxB)
B,(AxC)
I tried expanding everything by using A=(a1,a2,a3) B=(b1,b2,b3) C=(c1,c2,c3), but that actually led me to 0 for the initial statement... don't know, maybe I made a mistake somewhere, but I don't have time to work on it atm.
Ok, I figured out the first one... silly me:
A,(CxB)=4
because: (BxC)=(-CxB) and that one can be proved pretty easy by expanding ... maybe I just need to try the whole expansion thing on the others as well... hmm
ok, yes, through some expansion and regrouping, I get that C,(AxB) = A,(BxC) = -4 ....one more left to go!!!
and, as you guessed, B,(AxC) = -A,(BxC) = 4
sorry for the stupid question, and I hope I'm not expected now to write out the entire solution... but basically, you replace A, B, and C with the (x,y,z) variable I assigned earlier (or any others you want), perform the operations, and expand completely... then you just regroup expanded variables of each statement such that it resembles the initial statement (A,(BxC)) in its expanded form, and see if you have to add a negative sign or not for them to be equal...
Last edited by drak10687; September 13, 2007 at 11:34 PM.
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Ok this is probably easy for most, seeing as i am incompetent in math, especially algebra, so if anyone can help
i did a 30 parter test, seemed quite long to me
i got a c, and i can raise it to a B, my teacher is a hard grader..
solve the system of equations "algebraically". Show all of your work leading to the solution and explain how your solution would relate to the graph of the system
3x-2y=6
y=-1/2x+4
A) explain how you could explain to skippy, your skeptical classmate that you anwser is correct
B) find the x- and y- intercepts for the equation 3x-2y
---------------------
If f(x)=xsquarded-4x-5, find the following
A)F(-2) B) f(10) C) f(0)
-----------------------
and last would be the worse is
the quadratic formula:hmmm:
solve the equation using the quadratic forumla show work!
3xsquared+7x=neagtive12
Maybe or maybe not i may get this all done at school, i doubt it though, so i just wanted to see if anyone can anwser my 3 basic anwsers so my c and be a b, and have one less thing to worry about this school year.
I'll try to get back on twc later, after i finished all my other work and after csp.
thanks for taking your time :hmmm:
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Sgt. Timothy Prunty: End of Watch 10/24/10: Shreveport PD. Louisiana
Ok i really need help now
no one really up to this?
20,284 Officers Lost in the Line of Duty as of 2010-12 this month- 124 this year
Red: Suspect inflicted: Blue Accident
Officer Christopher A Wilson: End of Watch 10/27/10: San Diego PD, CA
Lt. Jose A Cordova Montaez: End of Watch 10/26/10: Pureto Rico PD
Cpt. George Green: End of Watch 10/26/10: Oklahoma Highway PD
Deputy Sheriff Odelle McDuffle Jr. 10/25/10: Liberty Country SD, Texas
Officer John Abraham: End of Watch 10/25/10: Teaneck PD New Jersey
Sgt. Timothy Prunty: End of Watch 10/24/10: Shreveport PD. Louisiana
A) Who the hell is skippy?
B)
3x-2y=6
y intercept = 0,y
x intercept = x,0
so...
Plug it on in there.
3(0)-2y=6
0-2y=6
-2y=6
y=6/-2
y=-3
y intercept: 0,-3
x intercept: 2,0
I wish I could help more, but don't remember anything else. You might want to wait for somebody to check the intercepts, too. Math has never been my strong point
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Ok, first question:
Just substitute y, which =-1/2x+4 into 3x-2y=6, so you get 3x-2(-.5x+4)=6. Then you solve for x. Then you take that x and plug it in to either of the equations to get the y. That's where the two lines intersect.
------------------
Just plug in those numbers into x, not too hard. Just grab a calculator and plug in (-2)^{2}-4(-2)-5 and (10)^{2}-4(10)-5 and (0)^{2}-4(0)-5
Not too hard
---------------
-b plus or minus the square root of b^{2}-4ac, all divided by 2a
3x^{2}+7x=-12. Set it equal to zero, so 3x^{2}+7x+12=0
a=3, b=7, c=12. Plug those number into the formula, and voila.
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I shouldn't be so willing to just give people the answers, I guess, but I'm feeling generous. If you can't solve most of that yourself, be aware that you're going to do very poorly in future math courses. You need to know this stuff to pass high school math.
Does that second equation have a term of x/2 or 1/(2x)? I'm assuming the former, since it's then linear like the other.
3x - 2y = 6
y = -x/2 + 4
therefore:
3x - 2(-x/2 + 4) = 6
3x + x - 8 = 6
4x = 14
x = 7/2
y = -(7/2)/2 + 4 = -7/4 + 16/4 = 9/4
The solution is the intersection of the two graphs.
Get a math whiz to agree with you. If that doesn't work, he's an idiot and he doesn't deserve to believe the right answer.
I'm assuming you meant 3x - 2y = 6. In that case, you can find the x- and y-intercepts by plugging in x = 0 and y = 0 respectively:
3(0) - 2y = 6
y = -3 is the x-intercept
3x - 2(0) = 6
x = 2 is the y-intercept
f(-2) = (-2)² - 4(-2) - 5 = 4 + 8 - 5 = 7
f(10) = (10)² - 4(10) - 5 = 100 - 40 - 5 = 55
f(0) = (0)² - 4(0) - 5 = -5
3x² + 7x = -12
3x² + 7x + 12 = 0
x = [ -7 +/- sqrt( 7² - 4(3)(12) ) ] / 2(3) = [ -7 +/- sqrt( 49 - 144 ) ] / 6 = -7/6 + i sqrt(95) or -7/6 - i sqrt(95)
Note that if you haven't covered imaginary numbers, the answer will be different:
x = [ -7 +/- sqrt( 7² - 4(3)(12) ) ] / 2(3) = [ -7 +/- sqrt( 49 - 144 ) ] / 6. 49 - 144 = -95 < 0, so there are no solutions.
Here's one I actually managed to solve, but I can't explain how.
A wagon is released with an acceleration of 0.5 m/s^2
Bob can run at 8m/s, how long can bob wait before he sets off after the wagon if he is to catch it?
Solving it graphically, piece of cake, just find the tangent on the graph 0.25t^2 which has the slope 8x, but the problem is to solve it mathematicaly.
I realized that if object A has an a=0.5 then object B, who has a constant speed, can always, no matter what its speed is, wait until A has reached half of B's speed before B has to set out after it. I also saw that the total time passed before they met would be the double of the time B could wait.
This gives the formula: 0.25t^2=x*(t-y) where both x and y is 1/2 t, so I can write it like this: 0.25t^2=1/2t*(t-1/2t)
You can all see that I end up with 0.25t^2=0.25t^2, so it is correct, but the problem is that this just popped into my head. I cannot explain how I came to think of those values of t,x and y, I didn't read it off the graph, it just came to me. I sometimes work ahead of myself.
I'm just wondering how I can solve this properly? While the teachers and co-students are sometimes impressed, they don't like to hear "I just saw the answer".
Last edited by Mathias; September 27, 2007 at 09:38 AM.
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The way I would do it would be thus: let us say we start the wagon at t = 0, and Bob at some later t = t_{0}. Let both of them start at x(0) = 0, and of course v(0) = 0. Since the wagon is under constant acceleration and Bob is under constant speed, we know their positions are (omitting units, and ignoring Bob's position before he actually starts running)
x_{W}(t) = (0.5)t²/2
x_{B}(t) = 8(t − t_{0})
When the two meet, by definition, their position must be equal, so we have
0.25t² − 8t + 8t_{0} = 0
This is a quadratic equation, so we know that the solutions will only exist if b² − 4ac >= 0, i.e., 64 − 8t_{0} >= 0, t_{0} <= 8. Checking my work, I find that if Bob waits 8 seconds, then after a further eight seconds he will have traveled 64 meters, and the wagon will also have traveled (0.25)(16)² = 64 meters, which is exactly as desired.
Your observation, that Bob can wait until the wagon has traveled half the distance he can travel per second, is thus actually wrong, unless you mistyped it. (I'm pretty sure I'm not the one misinterpreting "A has reached half of B's speed", but that could also be the issue. It's easier to check if you give your numerical solution.) At the maximum time he can wait, eight seconds, the wagon has moved (0.25)(8)² = 16 m, which is not half his speed but twice it.
It's true, however, that after he sets out, it will take him the same time to reach the wagon as the time he waited, if he waited the maximum possible time and the wagon's acceleration is exactly one-half in the units used. This is not difficult to see from the equations if you take his speed equal to v and get the somewhat more general solution to the time of collision
t = 2v ± 2 sqrt( v(v − t_{0}) )
which shows that the maximum possible time he can wait is t_{0} = v, in which case the time of collision is simply equal to 2v.
On the subject of kinematics, I've been wondering about one of the equations that we're given in our course, in this case for trajectory.
Purely theoretically (we won't need to apply this sort of thing), how would you solve the equation for θ, given that it is present twice?
I presume that there's an algebraic way to remove it; but we haven't been taught it
y = gx²/2u²cos²θ + x.tanθ
where
y = vertical displacement
g = gravity
x = horizontal displacement
u = velocity
θ = angle of elevation of velocity given
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There's no systematic way to do this sort of thing: it varies depending on what functions the desired variable appears in. Since these are trig functions, we will of course be using trig identities. With that in mind, this isn't hard:
sec² θ + a tan θ + b = 0
(tan² θ + 1) + a tan θ + b = 0
tan² θ + a tan θ + b + 1 = 0
tan θ = −0.5a ± 0.5sqrt( a² − 4b − 4 )
θ = arctan[ −0.5a ± 0.5sqrt( a² − 4b − 4 ) ]
You can consult the original equations to find that a = 2u²/gx and b = −2yu²/gx².
If you ask for solutions to an equation like x + e^{x} = 0, of course, you're probably out of luck. Numerical approximations are the only way if your functions aren't particularly simple or uniform. In this case we could use a Taylor series expansion to get an approximate solution: 1 + 2x + x²/2 = 0 gives x ≈ −2 ± sqrt(2) ≈ −0.586 or −3.414. The latter can quickly be discarded as impossible, and adding more terms will give a better estimation of the former (I picked quadratic so I could easily solve it). Actually −0.586 + e^{−0.586} = −0.0295 or so, which isn't bad. But you probably can't get a simple representation of the number. Or if you can, you certainly can't for something like (x^{6.235ix²} + e^{tanh(logφ x)})^{Γ(x)} − 3.2295 = 0.