SuicideFuel Math thread problem (official)

[trying to ascend]

Calculate how possible it is for me to get some bitches.

Infinitesimal, there is hope bhai

 

[Ahnfeltia]

Post problems for people to solve

For funsies?

 

[trying to ascend]

For funsies?

Yes

 

[Ahnfeltia]

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.

 

[Ahnfeltia]

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]

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!

 

[Ahnfeltia]

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.

 

[trying to ascend]

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]

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.

 

[trying to ascend]

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

 

[trying to ascend]

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.

 

[Intellau_Celistic]

fuck math

The current respondent is...a math graduate student, if it soothes your mind.

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Comprehensive encyclopedia of mathematics with 13,000 detailed entries. Continually updated, extensively illustrated, and with interactive examples.

mathworld.wolfram.com

 

[Intellau_Celistic]

Problems:

Mathematics Stack Exchange

Q&A for people studying math at any level and professionals in related fields

math.stackexchange.com

 

[trying to ascend]

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

 

[depressedblackcel]

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

 

[trying to ascend]

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

Aren't you the guy that wanted to learn programming?

 

[depressedblackcel]

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.

 

[trying to ascend]

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]

@Truckzo

Observe this thread closely. Time to develop.

 

[Truckzo]

AYO I NEED HELP


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

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

 

[Intellau_Celistic]

AYO I NEED HELP


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

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

Ch. 2 Introduction to Equations and Inequalities - Algebra and Trigonometry | OpenStax

For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some for...

openstax.org

 

[NowItsSlimeTime]

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!

 

[Truckzo]

You are in luck, I love algebra!

i got them myself lmao I just panicked

 

[Truckzo]

You are in luck, I love algebra!

ok wait wait wait could you awnser this

1. if f(x)=4-12 what is f(2)?

 

[NowItsSlimeTime]

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 =?

 

[NowItsSlimeTime]

ok wait wait wait could you awnser this

1. if f(x)=4-12 what is f(2)?

Oh yeah one sec

 

[NowItsSlimeTime]

Its 2

 

[Truckzo]

Its 2

the whole equation? is 2?

 

[NowItsSlimeTime]

Heres why:

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

Then 6 - 4 is 2

 

[NowItsSlimeTime]

the whole equation? is 2?

Nah just f

 

[Truckzo]

Nah just f

dude is the awnser 2 or are you just trolling me?


this is for a test man im cheating right now

 

[NowItsSlimeTime]

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.

 

[Truckzo]

o

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.

h my god oh mygod

 

[NowItsSlimeTime]

o

h my god oh mygod

Was it wrong

 

[Truckzo]

Was it wrong

idk idk im getting different awnsers im off to bed cya

 

[trying to ascend]

You are in luck, I love algebra!

Simplify the expression below:

((Sqrt of (5) + 2)^1/3 + (sqrt of (5) - 2)^1/3)^2014

 

[ilieknothing]

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]

Simplify the expression below:

((Sqrt of (5) + 2)^1/3 + (sqrt of (5) - 2)^1/3)^2014

Spoiler: answer

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.

 

[trying to ascend]

Spoiler: answer

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

 

[Intellau_Celistic]

Resource:

Math Is Fun Forum

 

[Pepecel]

1/0

 

[trying to ascend]

1/0

Undefined

 

[Pepecel]

Undefined

incredible math genius

 

[ChocolateSmores]

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]

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

 

[car+tree=good]

dog+foid+bed

car+tree=good?

 

[Linesnap99]

car+tree=good?

probably, other factors like (speed) needs to be added or multiplied.

 

[Filthy_Reject]

@trying to ascend what accounts for 80, but only wants 20

 

[trying to ascend]

The sum of the first n terms of an arithmetic progression is equal to 110 and the sum of the first 2n terms is equal to 420. Knowing that the first term is 2, what's the value of the sum of the first 3n terms?

 

[trying to ascend]

@trying to ascend what accounts for 80, but only wants 20