Linesnap99
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Post problems for people to solveWhat the fuck is the point of this thread?
Infinitesimal, there is hope bhaiCalculate how possible it is for me to get some bitches.
For funsies?Post problems for people to solve
YesFor funsies?
Compute the integral of 1/(1+x^5)
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.i could be wrong but i dont think there is anything more you can do with this.
Notice that the equations can be more appealingly rewritten as follows: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?
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.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)
Wrong, it's a vertical line, because such equality of hyperboles only holds for x = 1,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.
Correct!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.
I don't understand what you mean. I believe the hyperbolae I obtained are disjoint.Wrong, it's a vertical line, because such equality of hyperboles only holds for x = 1,
sqrt(x² + 2x + 1) + ∣ y - 2∣ = 1/2.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.
Rewriting yieldssqrt(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
CorrectRewriting 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.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.
fuck math
Aren't you the guy that wanted to learn programming?MOGS THE FUCK out of me. I'm pretty bad at math
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.Aren't you the guy that wanted to learn programming?
I mean, if you manage to get that score, you aren't pretty bad at all.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.
AYO I NEED HELP
I NEED Y=-6=3(X-4) GRAPHED RIGHT NOW
AND I NEED Y+4=-2(X-5)
You are in luck, I love algebra!AYO I NEED HELP
I NEED Y=-6=3(X-4) GRAPHED RIGHT NOW
AND I NEED Y+4=-2(X-5)
i got them myself lmao I just panickedYou are in luck, I love algebra!
ok wait wait wait could you awnser thisYou are in luck, I love algebra!
AYO I NEED HELP
I NEED Y=-6=3(X-4) GRAPHED RIGHT NOW
AND I NEED Y+4=-2(X-5)
Oh yeah one secok wait wait wait could you awnser this
- if f(x)=4-12 what is f(2)?
the whole equation? is 2?Its 2
Nah just fthe whole equation? is 2?
dude is the awnser 2 or are you just trolling me?Nah just f
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.dude is the awnser 2 or are you just trolling me?
this is for a test man im cheating right now
h my god oh mygodI think so yeah. I could be wrong, I am not the best at math, but to the best of my ability it is 2.
Was it wrongo
h my god oh mygod
idk idk im getting different awnsers im off to bed cyaWas it wrong
Simplify the expression below:You are in luck, I love algebra!
Simplify the expression below:
((Sqrt of (5) + 2)^1/3 + (sqrt of (5) - 2)^1/3)^2014
CorrectSeeing 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.
Undefined
incredible math geniusUndefined
Math le bad even doe its a beautiful language that uses symbols to manipulate our environment with the concept of abstractnessWhat the fuck is the point of this thread?
Not how I would describe math, but it's not entirely inaccurate I guess.Math le bad even doe its a beautiful language that uses symbols to manipulate our environment with the concept of abstractness