（x+2y)/2+(3x-y)/3=5 ① 0.2(x+2y)+0.8(3x-y)=4 ② Calculate this binary linear equation, and add money when friends answer it in a minute

（x+2y)/2+(3x-y)/3=5 ① 0.2(x+2y)+0.8(3x-y)=4 ② Calculate this binary linear equation, and add money when friends answer it in a minute

(x + 2Y) / 2 + (3x-y) / 3 = 5 ① multiply both sides of the equation by 6:3 (x + 2Y) + 2 (3x-y) = 30, and then expand and merge the same items into: 9x + 4Y = 30 0.2 (x + 2Y) + 0.8 (3x-y) = 4 ② multiply both sides of the equation by 10:2 (x + 2Y) + 8 (3x-y) = 40, and then expand and merge the same items into: 26x-4y = 40 ① + ② get: 35x = 7

2Y = - 1 + XY3 (the third power of Y). Take the derivative of X. the answer is 2Y '= the third power of Y + 3x times the second power of Y times y', for explanation. Thank you

Take the derivatives on both sides, and you get
(2y)'=(-1+xy^3)'
2y'=(xy^3)'
2y'=x'y^3+x(y^3)' (uv)'=u'v+uv'
2y'=y^3+x*3y^2*y'
2y'=y^3+3xy^2*y'

Find the first and second derivatives of the function y = y (x) determined by x = cost * e ^ t and y = Sint * e ^ t

dy/dt=e^t(cost+sint)
dx/dt=e^t(cost-sint)
So dy / DX = (dy / DT) / (DX / DT) = (cost + Sint) / (cost Sint) = 1 /) cos ² t-sin ² t)=1/cos2t=sec2t
d(dy/dx)/dt=(sec2t)'=2sec2t × tan2t
d ² y/dx ²= [d(dy/dx)/dt]/[dx/dt]=[2sec2t × tan2t]/[e^t(cost-sint)]

Calculate the total length of Starline x = ACOS ^ 3 (T), y = asin ^ 3 (T)? Draw a prismatic figure in a coordinate system. The range of X axis is - A to a, and the range of Y axis is - A to a, I think the range of T is 0 to pi / 2, and the final result is multiplied by 4, Does anyone know how to find 12a?

Indeed, just calculate the length of the first quadrant and multiply by 4
First, the arc differential DS = √ [(DX) ^ 2 + (Dy) ^ 2] = √ [(x ') ^ 2 + (y') ^ 2] DT = 3A | sintcast | DT, X ', y' denotes derivation
Secondly, the arc length s = 4 ∫ (0, π / 2) 3a | sintcostdt = 12a ∫ (0, π / 2) sintcostdt = 6A

Use Green's formula to find the area of star line x = ACOS ^ 3T, y = asin ^ 3T,

Using Green's formula to find the star line x = ACOs ³ t,y=asin ³ Area of T. s = (1 / 2) ∮ XDY YDX = [0,2 π] (1 / 2) ∫ (3a) ² cos⁴tsin ² t+3a ² sin⁴tcos ² t)dt=[0,2π](3a ²/ 2)∫(cos ² tsin ² t(cos ²...

Star curve, x = ACOS ^ 3T, y = asin ^ 3T, find the area enclosed by the curve? Can it be solved in polar coordinates? What other methods are there without ∫ YDX

Theoretically, it can be expressed in polar coordinates: P = a * (sin ^ 6T + cos ^ 6T) ^ (1 / 2), in the integral. Area s = P ^ 2 (T) DT (the upper and lower limits of the integral are 2pi, 0), but the integral is more complex