Yes, you can, but keeping (6n+7) on the denominator (and the minus sign)I need a derivitive power series, probably should've said that. So can't just do n(4n-7)*x^n-1
(so sum of -n(4n-7)/(6n+7) * x^(n-1) )
Or I don't understand the issue?
Yes, you can, but keeping (6n+7) on the denominator (and the minus sign)I need a derivitive power series, probably should've said that. So can't just do n(4n-7)*x^n-1
I need a derivitive power series, probably should've said that. So can't just do n(4n-7)*x^n-1
Trying to work backward, I save scummed the answer out of the multiple choices, and itr has a 6n+13 on the bottom. I figured that I have to differentiate, distribute the n, and then change all the ns in the equation to n+1s. Not sure why I have to do that
When you differentiate you get
-n*(4n-7)*x^(n-1)/(6n+7)
Now, a power series has a general term with the form a_n*x^n, not a_n*x^(n-1), so you need to do k = n - 1 and you get:
-(k + 1)*(4k - 3)*x^k/(6k+13)
Yes, you can, but keeping (6n+7) on the denominator (and the minus sign)
(so sum of -n(4n-7)/(6n+7) * x^(n-1) )
Or I don't understand the issue?
Not Math but close enough!
I need help with this, I'm having so much trouble seeing the angels with these type of questions
I'm not sure, is 57 supposed to be between between the red F line and the X axis or the the entire angle from the 37 of the mg to the F red line?
The 57 degrees is the angle (cf. Hot Fuzz) between the vector F and the local x-axis.
On the left picture, we see that F acts at an angle of 20 degrees; that, plus the angle of 37 degrees from inclination, gave us 57 degrees.
So I'm trying to prove this:
I figure I will have to prove by four cases, and for each prove the left hand side and then the right hand side.
If you have proved the triangle inequality already, it can be used to quickly do your current proof:
http://www.mathwords.com/t/triangle_inequality_with_absolute_value.htm
If you haven't, you may want to just prove that first so you can use it.
Thanks. So I've proven the triangle inequality and that made it easier to solve the problem without using proof by case.
But let's say I was asked to solve it by case on an exam. If I were to solve this problem by cases, could I first add +3|y| to both sides of the equation and still preserve the inequality to prove it or would this be considered a completely different proof?
I haven't finished all of the cases yet, but does this look valid?
Is there a good resource that can explain(not teach) solving systems with matrices?
My third eye wont open and I just don't get it 100%.
By "solve" I mean get to eshelon form.
I've had two more tests since the last time I posted and have kept up the good grades! I got a 97 on the second one and 93 on the most recent! I thought I was going to bomb the most recent one but to my surprise, I didn't! We're now on Law of Sines and Cosines and started some calculator stuff today and this seems easier than the previous stuff. I should probably start looking back at some stuff to prepare for the cumulative final.
This maybe? Not really sure what difference there is between explaining and teaching.
Also what is the exact part you don't get? Maybe someone here could try to explain it.
so to find the number of hands in poker with 2 pairs, the equation is:
nCr(13,2) * nCr(4,2) * nCr(4,2) * nCr(44,1)
In the first term you choose 2 out of 13 different card denominations in the same step. What I don't understand is, why can't I choose one denomination for a pair, then choose another in 2 separate steps using this equation:
nCr(13,1) * nCr(4,2) * nCr(12,1) * nCr(4,2) * nCr(44,1)
The 2nd equation is twice that of the first, but I'm not seeing where the 2x comes from...?
Fairly simple question but I'm having trouble with it as I haven't done math in a while.
((x^(4a))/(y^(2a)))^(3b^2)
Need to double check on statistics problem. Its very simple but I feel like I missed an obvious thing.
If part 1 has an .85 reliability and part 2 .93, what is the probability the system will fail (both parts need to operate to be successful).
Is it as simple as multiplying them together and taking the inverse? Or is this a nonmutually exclusive event?
Oh whoops, I just need to evaluate what I posted.Where's the question?
Can you write the exact wording of the question? Evaluate doesn't really work as a question for this.
Alright, so the question is about expanding (which is, to get rid of the parentheses). The goal is simply to get used to the rules relating to exponents in algebra.
The first rule you want to use is (a/b)^m = a^m/b^m. Applying the rule, you get this :
(x^(4a))^(3b^2)
______________
(y^(2a))^(3b^2)
Next rule you'll want to use is (a^m)^n = a^(mn). Basically, multiply the exponents together.
After that, the problem is over because the bases (x and y) are different, so there's no simplication possible.
Unless you are given BD = BX (where X is the point on the radius of the circle that intersects line BC), I do not believe you have any mathematical basis to assume that point B is the center of the segment you observe -- but by visual inspection it appears that it is, as you surmised.
Unless you are given BD = BX (where X is the point on the radius of the circle that intersects line BC), I do not believe you have any mathematical basis to assume that point B is the center of the segment you observe -- but by visual inspection it appears that it is, as you surmised.
BD=BX doesn't seem like it would imply that B is the center. If you were to form similar triangles by shifting B back and forth along the line bisecting angle B, you would still have BD=BX in any of the resulting diagrams.
I think the intention was to have it appear that the curve meets each line at a 90 degree angle, from which you could infer that B is the center.
Sure. I was assuming the curve was part of a circle as in the original post, and that the only question was where the center lies. If you assume nothing at all about the curve then the the problem is clearly nonsensical.Here is an ellipsoid [I guess more properly an ellipse, since it is in two dimensions] that has the property that the arc meets each line forming a 90 degree angle, but where BD and BX are not equal; by visual inspection, is the arc a quarter-circle, or a stretched ellipsoid?
(But it is equally trivial that my condition was insufficient; observe that the red and blue arcs clearly do not form a quarter-circle but BD = BX
When doing a partial differentiation with respect to x, just deal with y as if it was a constant.Does anyone understand partial differential equations at all? I have no fucking clue how to go about solving them and all these greek letters are just confusing me.
Like, I gotta solve du/dx + 3u = xy^2 + y and I just don't even know where to begin.
That's the thing though, I don't. It's got du(x,y)/dx AND u(x,y) terms so I cant work out how to separate the variables; or if i want one of them substitutions with Greek letters or what.When doing a partial differentiation with respect to x, just deal with y as if it was a constant.
For example,
d/dx( 3.x^2 + 4.y + x.y^2 + sin(x.y) )
is simply
6.x + 0 + y^2 + y.cos(x.y)
If you know how to solve du/dx + 3u = x.4^2 + 4, it's the same. Just treat y as if it was a constant.
That's the thing though, I don't. It's got du(x,y)/dx AND u(x,y) terms so I cant work out how to separate the variables; or if i want one of them substitutions with Greek letters or what.
Those images are rather small but I guess the first one is (1/3)^(1/2), right? What do you get as a result and how you get it?
so, as usual, you begin without the right part:
du/dx + 3u = 0
means u(x, y) = A exp(-3x)
A can have a dependancy in y, since y is a constant when differenciating with respect to x (just substitute if you need to check)
Then, you look for a specific solution to du/dx + 3u = xy^2 + y
You look for an affine solution in x, so B x + C
Let's substitute, you get B + 3 B x + 3 C = xy^2 + y
By identification, 3 B = y^2, so B = (1/3).y^2
And B + 3 C = y, so C = (1/3).(y - (1/3).y^2)
This the solutions are f(x, y) = A exp(-3x) + (1/3).x.y^2 + (1/3).(y - (1/3).y^2)
(unless I've made a mistake, it's hard typing this kind of stuff)
A can be ANY function of y: e.g. 2, 14.y, sin or cos.exp(-y)
Now those are easier to read. That second one is a pain to count in the head lol. I try to write detailed answers.
Done. Will see if I can get a good image of it. As for advices before giving you straight answers, change the roots into exponents for easier calculation. Just need to remember the rules on how to calculate them. This will work well with the second and third problem. First problem I kinda want to see what you have done since you have probably done it right and the problem is just with presentation.
For the first problem I got √(1/3) with both the numerator and denominator being raised to the 3rd power and subsequently getting √1/27. Who knows if that's right though