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[ILikePie]

It should be, but I keep getting the wrong answer.

[Darvi]

Aaaaaactually it's a first degree equation with a parametre I just noticed.

[Jim Groovester]

It should be, but I keep getting the wrong answer.

What's the right answer and what answer do you keep on getting?

[Darvi]

Domain:x!=0, m!=0, x!=2-4m

1/mx = 4m/(x + 4m - 2)
<=>x+4m-2=4m^2x
<=>(4m^2-1)x=4m-2
<=>m=0.5 or x=(4m-2)/(4m^2-1)
<=>m=0.5 or x=2(2m-1)/(2m+1)(2m-1)
<=>m=0.5 or x=2/(m+1) and m!=-1 or 0.5

Should be about right.

And now I need painkillers.

[Another]

Domain:x!=0, m!=0, x!=2+4m

1/mx = 4m/(x + 4m - 2)
<=>x+4m-2=4m^2x
<=>(4m^2-1)x=4m-2
<=>m=0.5 or x=(4m-2)/(4m^2-1)
<=>m=0.5 or x=2(2m-1)/(2m+1)(2m-1)
<=>m=0.5 or x=2/(m+1) and m!=-1 or 0.5

Should be about right.

And now I need painkillers.

* x!=2-4m in domain definition and
* x=2/(2m+1) m!=-0.5 in the last line.

[Darvi]

I ninja-edited the first part and the second part is kinda there already.

Makes me kinda wonder though how it took you 5 minutes to post that

[ILikePie]

I see. Thanks, my mistake was going from 4m^2 - 1 to (4m - 1)(4m + 1).

[ed boy]

-dice question-

Let's have another go with this, this time using probability generating functions.

Spoiler: maths (click to show/hide)

If you don't know, a probability generating function tells you how a discrete random variable is distributed. It is a function of some variable (we will use S) such that for a random variable X;
G_X(s)=E(s^X)
G_X(s)=Σ_ip(X=i)sⁱ
where i ranges from  to infinity. If we say that a single dice X has k sides, with a p=k^-1 chance for each one, then we have:
G_X(s)=Σ_ipsⁱ
G_X(s)=pΣ_isⁱ
where i ranges from 1 to k.

Now suppose we have N such dice, where each roll is independent of each other, and identically distributed. If Y is the total of these N dice, then we have:
G_Y(s)=E(s^Y)
G_Y(s)=E(s^{X₁+X₂+...+X_N})
G_Y(s)=E(s^X₁s^X₂...s^X_N)
G_Y(s)=E(s^X₁)E(s^X₂)...E(s^X_N)
G_Y(s)=G_X(s)^N
G_Y(s)=p^N(Σ_isⁱ)^N

Note that this is a polynomial in terms of s. If you are using dice of different numbers of sides, then find the polynomial for each one, then multiple them together to get a new polymonial.

If we want to get the probability of some result R, then we just need to find the coefficient of s^R. If I can find an easy way of doing this, I'll add it.

[ILikePie]

I keep doing something stupid but I don't what.
"-4 / (a + 2)" is the tip of a parabola, and I need to find the values of "a" where the tip is to the left of the line "x = 3".

I've got this:

Code: [Select]

-4 / (a + 2) < 3   / : -4
a + 2 > 3/-4       / +  2
a > 3/-4 - 2

a > -2.75
(and a != -2)
The answer in the book is either a > -2 or a < -3.333. What do I keep missing?

[Jim Groovester]

Why are you dividing by -4, but not keeping (a + 2) in the denominator?

[ed boy]

I'm confused. If you have a parabola, then any point on the parabola will have two co-ordinates - the x-coordinate and the y-coordinate. You've only given us one: -4 / (a + 2)
If you mean (-4,a+2) as the two coordinates, then the condition "to the left of x=3" is always satisfied for all values of a.
If you mean ^-4/_(a+2) is the y-coordinate, then for each value of a, x can take any value, some of which will be satisfactory and some of which will not.
If you mean ^-4/_(a+2) is the x-coordinate, then we have:
^-4/_(a+2)<3
We there are two possibilities: (a+2)>0, or (a+2)<0. If (a+2)>0, then:
-4<3(a+2)
-4<3a+6
-10<3a
a>-¹⁰/₃
However, this is when a+2>0, so a>-2
However, a>-¹⁰/₃ is implied by a>-2, so we can take the condition a>-2.
Now consider when (a+2)<0, which is equivalent to a<-2.
-4>3(a+2)
-4>3a+6
-10>3a
a<-¹⁰/₃
so in this situation, we have that a<-2 and a<-¹⁰/₃ both hold.
The latter implies the former, so we can take the single condition a<-¹⁰/₃.
So a<-¹⁰/₃ or a>-2.

[ILikePie]

sin a = -5/13 and 270 < a < 360. How do I, without using a calculator, find sin(90 - a).

[ed boy]

you use compund angle formulae.

sin(a+b)=sin(a)cos(b)+sin(b)cos(a)
sin(90-a)=sin(90)cos(-a)+sin(-a)cos(90)
sin(90-a)=cos(a)

now you need to find cos(a). You can use:
sin²(a)+cos²(a)=1
cos²(a)=1-sin²(a)
cos²(a)=1-(⁵/₁₃)²=¹⁴⁴/₁₆₉
you know that 270<a<360, so you know that cos(a) is positive.
so cos(a)=¹²/₁₃

[ILikePie]

If I've got a right triangle (ABC) where B = 90 degrees, and I'm told the coordinates of B are (10,8), and A's are (2,4). How do I get C's coordinates if I know it sits somewhere on the x axis (i.e. C(x,0) )?

[Darvi]

Since AB and BC are perpendicular, either it's X_BA=k*Y_BC and Y_BA=-k*X_BC, or X_BA=-k*Y_BC and Y_BA=k*X_BC,which are the coordinates of the vectors BA and BC respectively. (k is positive, btw)

Now, since C is to the below B (Y_BC is negative) and A is to the left of B(X_BA is negative), we come to the conclusion that the second must be true.

Bluh, I suck at reading notations. Fixed.