Nana to Kaoru - Step Up "S&M" Love Comedy - Vol. 10 Ch. 79 - An Infinite Gap
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oohf man i felt that way to hard.
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The empathy I have for Kaoru makes this manga feel like a sardonic parody of my time through high school.
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Dang, I didn’t realize when I started this manga I was actually kinda like this mc
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MC in fuckin Japanese romcom mode.
Girl ok with being almost naked in front of him, kisses him, lets him do sexy and absolute kinky shit with her and touch her boobs.
This dumbass: Ohh... maybe she likes me? No-no, it's impossible! Nooo, don't speak to me! Don't come close to me!
I bet Nana could fuck his brain out and ask him to marry her and he still would be like: No! There is no way she likes me.
Girl ok with being almost naked in front of him, kisses him, lets him do sexy and absolute kinky shit with her and touch her boobs.
This dumbass: Ohh... maybe she likes me? No-no, it's impossible! Nooo, don't speak to me! Don't come close to me!
I bet Nana could fuck his brain out and ask him to marry her and he still would be like: No! There is no way she likes me.
If anyone is curious about that one visible problem here is a solution.
We have some algebraic curve defined by its points (x,y) satisfying x²+2xy+5y²+2x+4y-5=0 (E). We seek the maximum value of x for a point on this curve. It is not explicitly said on which field (or ring) we are working, but it is rather natural to assume they are real numbers.
First let's decouple x from the other variable by defining z=y+ax, and substituting it in (E). We notice that the term with mixed variables vanishes for a=⅕.
We now get a new equation: 4x²+6x+25z²+20z-25=0.
Defining f:x↦4x²+6x and g:z↦25z²+20z-25, this boils down to f(x)+g(z)=0. They are two quadratic polynomial functions, and we notice that f has 0 and -3/2 as roots. Further for x=0, the equation has a solution (in fact, two) as there is a value of z that satisfies 5z²+4z-5=0 (a common factor of 5 was simplified). This can be seen from computing the discriminant Δ=4²-4*5*(-5)=116, which is positive.
That is, there is a point on the curve with x=0; meaning the sought-after maximum x is at least 0. But larger values of x are where f is increasing, so they correspond to smaller values of g(z). The function g has a minimum, and it is reached for z=-⅖. Plugging this in, the maximal x satisfies 4x²+6x-29=0. This has two solutions, and obviously the maximal x is the bigger one. After using and simplifying the quadratic formula, we have a maximal value of x of (-3+5√5)/4.
Remark that the solution being irrational, the problem would not have had an answer on rational numbers.
The corresponding point on the curve has for (x,y) coordinates ( (-3+5√5)/4 , -(1+√5)/4 ). We notice that y=-φ/2, with φ the (much too) lauded golden ratio, because the problem setter was being cute.
We have some algebraic curve defined by its points (x,y) satisfying x²+2xy+5y²+2x+4y-5=0 (E). We seek the maximum value of x for a point on this curve. It is not explicitly said on which field (or ring) we are working, but it is rather natural to assume they are real numbers.
First let's decouple x from the other variable by defining z=y+ax, and substituting it in (E). We notice that the term with mixed variables vanishes for a=⅕.
We now get a new equation: 4x²+6x+25z²+20z-25=0.
Defining f:x↦4x²+6x and g:z↦25z²+20z-25, this boils down to f(x)+g(z)=0. They are two quadratic polynomial functions, and we notice that f has 0 and -3/2 as roots. Further for x=0, the equation has a solution (in fact, two) as there is a value of z that satisfies 5z²+4z-5=0 (a common factor of 5 was simplified). This can be seen from computing the discriminant Δ=4²-4*5*(-5)=116, which is positive.
That is, there is a point on the curve with x=0; meaning the sought-after maximum x is at least 0. But larger values of x are where f is increasing, so they correspond to smaller values of g(z). The function g has a minimum, and it is reached for z=-⅖. Plugging this in, the maximal x satisfies 4x²+6x-29=0. This has two solutions, and obviously the maximal x is the bigger one. After using and simplifying the quadratic formula, we have a maximal value of x of (-3+5√5)/4.
Remark that the solution being irrational, the problem would not have had an answer on rational numbers.
The corresponding point on the curve has for (x,y) coordinates ( (-3+5√5)/4 , -(1+√5)/4 ). We notice that y=-φ/2, with φ the (much too) lauded golden ratio, because the problem setter was being cute.
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