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Registered: | February 27, 2024 |
Last post: | July 22, 2025 at 12:34 AM |
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Nope i was just giving you the question as you seemed confused about the answer key
this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this
Actually the question says (1,-2) does satisfy from that we find the constant, and then we have to find another point that satisfies, from the given options
Hello smthlikeyou11 <3 :))
No thanks a lot nexus, i appereciate it a lot <3 Btw do you know you have the most disliked post on the internet?
Your solution is wrong
I was actually just solving a previous year questions of this chapter, teacher didnt give us the question.
this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this
this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this
yes this is the whole ques
if a curve passes through the point (1,-2) and has slope of the tangent at any point (x,y) on it as (x²-2y)÷x, then the curve also passes through the point: (1) (3,0) (2) (√3,0) (3) (-1,2) (4) (-√2,1)
Now slope means differentiation and if we integrate the slope we get the curve and to find a constant we make use of the point given. But how do i integrate this
i have no idea what any of this means, can you please explain? We havent been taught integration yet so idk what an integrating factor is, i only know basic integration rules that are used in physics for taking elements
yes we havent been taught it yet, we are on AOD rn can you give the full solution
I need help with the integration of this dy/dx = x - 2y/x
i tried to take help from ai but this is what ir proved 💀
The correct answer is (3) (-1,2).
Given the slope of the tangent at any point (x, y) on the curve as (x²-2y)÷x, we can write the differential equation:
dy/dx = (x²-2y)/x
We are also given that the curve passes through the point (1, -2). We can use this information to find the equation of the curve.
Separating the variables, we get:
∫dy = ∫(x²-2y)/x dx
Integrating both sides, we get:
y = (1/3)x³ - 2x + C
Using the point (1, -2), we can find the value of C:
-2 = (1/3)(1)³ - 2(1) + C
C = -4/3
So, the equation of the curve is:
y = (1/3)x³ - 2x - 4/3
Now, we can check if the curve passes through the point (-1, 2):
2 = (1/3)(-1)³ - 2(-1) - 4/3
2 = -1/3 + 2 - 4/3
2 = 2/3 (which is true)
Therefore, the curve also passes through the point (-1, 2).
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Murder for the smart who pretend to be stupid, there is no need to pretend stupid in a debate unless they are trolling
But to destroy the sins, Kru would need Sha(h)zam uk
As an indian i really dont know why we are third
Then it wont be a debate, rather a murder
"Never argue with stupid people, they will drag you down to their level and then beat you with experience"
Im great too
Your name is butterfly youdk what a cocoon is??
Meanwhile me surviving on 2 chapattis everyday 💀
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