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SuicideFuel Math thread problem (official)

Linesnap99

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Ahnfeltia

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What the fuck is the point of this thread?
 
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Calculate how possible it is for me to get some bitches.
 
Ahnfeltia

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Compute the integral of 1/(1+x^5)
i could be wrong but i dont think there is anything more you can do with this.
Wolfram Mathematica disagrees. Curiously the solution features the golden ratio prominently. I imagine the way to solve for the antiderivative of 1/(1+x^5) is to notice that 1+x^5 = (1+x)(1-x+x^2-x^3+x^4) and to further notice that 1-x+x^2-x^3+x^4 = (1+ax+x^2)(1+bx+x^2) for some a & b by the fundamental theorem of algebra and judicious guessing. Solving for a & b readily yields that -- without loss of generality -- a is minus the golden ratio and b is the reciprocal of the golden ratio. At this point, you can probably go thru some onerous partial fraction decomposition to arrive at the answer.
 

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Ahnfeltia

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Consider a point P whose coordinates (x,y), x,y∈R satisfy the system

4 cossec(α)x − 6cotg(α)y = 4sen(a)

12 cossec(α)y − 8cotg(α)x = 0

where α is an angle in radians other than kπ (k∈Z). The locus described by the points P, as the angle α is varied, is a segment of?
Notice that the equations can be more appealingly rewritten as follows:

(csc α)(2x) - (cot α)(3y) = 4(sin α)

(csc α)(3y) - (cot α)(2x) = 0

Since (a^2-b^2)(c^2-d^2) = (ac-bd)^2-(ad-bc)^2 (a variant of the Brahmagupta-Fibonacci identity) and (csc α)^2 - (cot α)^2 = 1 (note that csc α and cot α are well-defined because α is not an integral multiple of π) it readily follows that 4x^2-9y^2 = 4(sin α)^2. Since α is not an integral multiple of π, the loci are nondegenerate hyperbolae.
Let a be a positive real number. Set S(a) to the value of the enclosed area bounded by the y-axis, by the parabola =x² and by the tangent line to the same parabola at the point (a, a²).


Find the limit: lim→+∞ S(a)/(a³ + a² + a + 1)
The tangent line to the parabola at (a,a^2) is readily seen to be y = 2ax-a^2 (use that the slope of the line is the derivative of x^2 at x=a). Ergo, we want to calculate the integral from x=0 to a of x^2-(2ax-a^2) = x^2-2ax+a^2 = (x-a)^2, which is easily found to be a^3/3. The desired limit therefore equals 1/3.
 
trying to ascend

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Notice that the equations can be more appealingly rewritten as follows:

(csc α)(2x) - (cot α)(3y) = 4(sin α)

(csc α)(3y) - (cot α)(2x) = 0

Since (a^2-b^2)(c^2-d^2) = (ac-bd)^2-(ad-bc)^2 (a variant of the Brahmagupta-Fibonacci identity) and (csc α)^2 - (cot α)^2 = 1 (note that csc α and cot α are well-defined because α is not an integral multiple of π) it readily follows that 4x^2-9y^2 = 4(sin α)^2. Since α is not an integral multiple of π, the loci are nondegenerate hyperbolae.
Wrong, it's a vertical line, because such equality of hyperboles only holds for x = 1,


The tangent line to the parabola at (a,a^2) is readily seen to be y = 2ax-a^2 (use that the slope of the line is the derivative of x^2 at x=a). Ergo, we want to calculate the integral from x=0 to a of x^2-(2ax-a^2) = x^2-2ax+a^2 = (x-a)^2, which is easily found to be a^3/3. The desired limit therefore equals 1/3.
Correct! :feelsokman:
 
Ahnfeltia

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Wrong, it's a vertical line, because such equality of hyperboles only holds for x = 1,
I don't understand what you mean. I believe the hyperbolae I obtained are disjoint.

PS A hyperbole is an exaggerated figure of speech. The conic section is written with an a at the end.
 
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I don't understand what you mean. I believe the hyperbolae I obtained are disjoint.

PS A hyperbole is an exaggerated figure of speech. The conic section is written with an a at the end.
sqrt(x² + 2x + 1) + ∣ y - 2∣ = 1/2.

x² + 36y² = 2(kx + 72y) - 140 - k²


Find all values of K, such that the system of equations above has only 1 solution
 
Ahnfeltia

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sqrt(x² + 2x + 1) + ∣ y - 2∣ = 1/2.

x² + 36y² = 2(kx + 72y) - 140 - k²


Find all values of K, such that the system of equations above has only 1 solution
Rewriting yields

|x+1| + |y-2| = 1/2

(x-k)^2 + 36(y-2)^2 = 4

which geometrically represent a diamond centered at (-1,2) and an (axis-aligned) ellipse centered at (k,2). Thus, in order for there to be only one solution, the rightmost point of the diamond and the leftmost point of the ellipse have to coincide (or vice versa). The rightmost point of the diamond has x-coordinate x = -1/2. The leftmost point of the ellipse will therefore have x-coordinate equal to -1/2 and y-coordinate equal to 2, so (k+1/2)^2 = 4. I.e., k = -1/2 + 2 (k = -1/2 - 2 would make x = -1/2 the rightmost point of the ellipse). The vice versa case similarly yields k = -3/2 - 2. Ergo, k is either 1.5 or -3.5.
 
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Rewriting yields

|x+1| + |y-2| = 1/2

(x-k)^2 + 36(y-2)^2 = 4

which geometrically represent a diamond centered at (-1,2) and an (axis-aligned) ellipse centered at (k,2). Thus, in order for there to be only one solution, the rightmost point of the diamond and the leftmost point of the ellipse have to coincide (or vice versa). The rightmost point of the diamond has x-coordinate x = -1/2. The leftmost point of the ellipse will therefore have x-coordinate equal to -1/2 and y-coordinate equal to 2, so (k+1/2)^2 = 4. I.e., k = -1/2 + 2 (k = -1/2 - 2 would make x = -1/2 the rightmost point of the ellipse). The vice versa case similarly yields k = -3/2 - 2. Ergo, k is either 1.5 or -3.5.
Correct :feelsokman:
 
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Rewriting yields

|x+1| + |y-2| = 1/2

(x-k)^2 + 36(y-2)^2 = 4

which geometrically represent a diamond centered at (-1,2) and an (axis-aligned) ellipse centered at (k,2). Thus, in order for there to be only one solution, the rightmost point of the diamond and the leftmost point of the ellipse have to coincide (or vice versa). The rightmost point of the diamond has x-coordinate x = -1/2. The leftmost point of the ellipse will therefore have x-coordinate equal to -1/2 and y-coordinate equal to 2, so (k+1/2)^2 = 4. I.e., k = -1/2 + 2 (k = -1/2 - 2 would make x = -1/2 the rightmost point of the ellipse). The vice versa case similarly yields k = -3/2 - 2. Ergo, k is either 1.5 or -3.5.
Be ABC an acute triangle, such that AH = AO, where H and O are. respectively, its orthocenter and circumcenter, inside the triangle and not belonging to any side. Calculate the angle HBO, knowing that BAH = 2HAO.

Triangle
 
Intellau_Celistic

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Problems:

 
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If x = cubrt(25) + cubrt(20) + cubrt(16), then x - 1/x² is equal to?
 
depressedblackcel

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MOGS THE FUCK out of me. I'm pretty bad at math
 
depressedblackcel

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Aren't you the guy that wanted to learn programming?
yes, I've taken physics 1 and 2, I did pretty good and got A's, I took pre-calc got an A but I put in soooooo much effort. I'm legit bad at math because I have to compensate far more than anyone does. Lets say you study 5 hours to get a A, I'll study 20 hours to get that A.
 
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yes, I've taken physics 1 and 2, I did pretty good and got A's, I took pre-calc got an A but I put in soooooo much effort. I'm legit bad at math because I have to compensate far more than anyone does. Lets say you study 5 hours to get a A, I'll study 20 hours to get that A.
I mean, if you manage to get that score, you aren't pretty bad at all.

Since many people, despite putting their maximum effort, do far worse than you
 
Intellau_Celistic

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@Truckzo

Observe this thread closely. Time to develop.
 
Truckzo

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AYO I NEED HELP


I NEED Y=-6=3(X-4) GRAPHED RIGHT NOW

AND I NEED Y+4=-2(X-5)
 
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AYO I NEED HELP


I NEED Y=-6=3(X-4) GRAPHED RIGHT NOW

AND I NEED Y+4=-2(X-5)

Second one is Y = 10X + 4 I believe.

Is the sexond one meant to have two =?
 
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NowItsSlimeTime

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Heres why:

2 divided on both sides is 6, because pemdas means we divide

Then 6 - 4 is 2
 
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dude is the awnser 2 or are you just trolling me?


this is for a test man im cheating right now
I think so yeah. I could be wrong, I am not the best at math, but to the best of my ability it is 2.
 
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I am rusty af, I do all my maths on a spreadsheet or I use R while wageslaving. I could participate if I was still in uni
 
Ahnfeltia

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Simplify the expression below:

((Sqrt of (5) + 2)^1/3 + (sqrt of (5) - 2)^1/3)^2014
Seeing the sqrt(5) reminded me of the golden ratio (denote it by q for convenience). Conveniently, q^3 = sqrt(5)+2 and 1/q^3 = sqrt(5)-2. Ergo, our expression becomes (q+1/q)^2014 = sqrt(5)^2014 = 5^1007.
 
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Seeing the sqrt(5) reminded me of the golden ratio (denote it by q for convenience). Conveniently, q^3 = sqrt(5)+2 and 1/q^3 = sqrt(5)-2. Ergo, our expression becomes (q+1/q)^2014 = sqrt(5)^2014 = 5^1007.
Correct :feelsokman:
 
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ChocolateSmores

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What the fuck is the point of this thread?
Math le bad even doe its a beautiful language that uses symbols to manipulate our environment with the concept of abstractness
 
Ahnfeltia

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Math le bad even doe its a beautiful language that uses symbols to manipulate our environment with the concept of abstractness
Not how I would describe math, but it's not entirely inaccurate I guess.

Also: welcome brocel
 

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