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SuicideFuel Problem

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Oldcel KHHV
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Consider the real function f, defined by f(x) = - 2/x and two circumferences C1 and C2, with centers in the origin. It's known that C1 is tangent to the graphic of F, and that the point of X = -1/2 belongs to both C2 and the graphic. Given that, what's the size of the annulus defined by C1 and C2?
 
Consider the real function f, defined by f(x) = - 2/x and two circumferences C1 and C2, with centers in the origin. It's known that C1 is tangent to the graphic of F, and that the point of X = -1/2 belongs to both C2 and the graphic. Given that, what's the size of the annulus defined by C1 and C2?
it's a scandal that we don't learn how to do maths in english at school. :feelsUgh:
 
Is this high school or college problem?
 
Vc ainda tá no ensino médio?
 
Consider the real function f, defined by f(x) = - 2/x and two circumferences C1 and C2, with centers in the origin. It's known that C1 is tangent to the graphic of F, and that the point of X = -1/2 belongs to both C2 and the graphic. Given that, what's the size of the annulus defined by C1 and C2?
What the fuck is this ? The answer is (-1/2)x2xpi

Your question seems weird and out
 
The answer is pi number
 
I probably didn’t understand the question lol
You have two circles, both with centers in the origin, and a function (x) -2/x

The function is tangent to one circle and it touches the other circumference when X = -1/2.

Which part didn't you understand?
 
Last edited:
You have two circles, both with centers in the origin, and a function (-2/x).

The function is tangent to one circle and it touches the other circumference when X = -1/2.

Which part didn't you understand?
Answer is 7pi
 
my iq is too low for this
 

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