Трикутник ABC задано координатами його вершин A(-1;0) B(1;2), C(-3;2) знайдіть кутC​

Verse 1
A A5
It seems the more we talk
Asus4 A
The less I have to say
F#m D5
Let's put our differences aside
A A5
I wanted to make you proud
Asus4 A
But I just got in your way
F#m D5
I found a place that I can hide

Pre-Chorus
E
Now everything is changing
D5 A
But I still feel the same
E
Are we running out of time

Chorus
A
What do I have to do
D5
To try to make you see
E
That this is who I am
D
And it's all that I can be

Verse 2
A A5
I tried to find myself
Asus4 A
Looking inside your eyes
F#m D5
You were all that I wanted to be
A A5
There must be something else
Asus4 A
Behind all the lies
F#m D5
That you wanted me to believe

Pre-Chorus
E
Now everything is changing
D5 A
But I still feel the same
E
Are we running out of time

Chorus
A
What do I have to do
D5
To try to make you see
E
That this is who I am
D
And it's all that I can be

A
What do I have to do
D5
To try to make you see
E
Trying to be like you
D A E
Isn't good enough for me

Bridge
D A E
And I won't let you go
D A E
And I won't let you down
D A E
I won
Примерно вот так;) по -крайней мере я так бы играла) Поблагодари меня, если не сложно)

А) x^3 + y^3 - 6xy = 0

Производная неявно заданной функции.

3x^2 + 3y^2*y' - 6y - 6x*y' = 0

Делим всё на 3

x^2 + y^2*y' - 2y - 2x*y' = 0

y'*(y^2 - 2x) = - x^2 + 2y

y' = (-x^2 + 2y) / (y^2 - 2x)

Б) y = (sin x)^(5x/2)

Производная такой функции равна сумме производных от степенной и от показательной функции.

y = f(x)^g(x)

y' = g*f^(g-1) *f'(x) + f^g*ln |f|*g'(x)

В нашем случае f = sin x; g = 5x/2.

y' = (5x/2)*(sin x)^(3x/2)*(cos x) + (sin x)^(5x/2)*(ln |sin x|)*5/2

В) x = √(2t - t^2); y = (1-t)^(-2/3)

y'(x) = dy/dx = (dy/dt) : (dx/dt)

dx/dt = (-2t+2) / [2√(2t-t^2)] = (-t+1) / √(2t-t^2)

dy/dt = -(-2/3)*(1-t)^(-5/3) = (2/3) / (1-t)^(5/3)

dy/dx = [(2/3) / (1-t)^(5/3)] : [(-t+1) / √(2t-t^2)] =

= [(2/3)*√(2t-t^2)] / [(1-t)^(5/3)*(1-t)] = [2/3*√(2t-t^2)] / [(1-t)^(8/3)]