WebFirst, suppose x−y is orthogonal to x+y. Then 0 = hx−y,x+yi = (x−y)T(x+y) = xTx+xTy −yTx−yTx = hx,xi+hx,yi−hy,xi−iy,yi = hx,xi−hy,yi since hx,yi = hy,xi. In other words, 0 = kxk2 −kyk2, so kxk2 = kyk2. Since the norm of a vector can never be negative, this implies that kxk = kyk. Thus, we see that if x−y is orthogonal to x+y ... WebI got: 4x −y = 22 Explanation: We can use the general expression for a line through a point of coordinates (x0,y0) and slope m ... 5x-y=3 Geometric figure: Straight Line Slope = 5 x …

Solve y^3 + y

Web2. If X and Y are independent, then E(XY) = E(X)E(Y). However, the converse is not generally true: it is possible for E(XY) = E(X)E(Y) even though X and Y are dependent. Probability as an Expectation Let A be any event. We can write P(A) as an expectation, as follows. Define the indicator function: I A = (1 if event A occurs, 0 otherwise. Then I WebSolve y' = y3 for y (0) = 1. This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer Question: … bitcoin alpha数据集

consider the initial value problem $ y

Webyn+1(x)=y 0 + Zx x0 f(t,yn(t))dt produces a sequence of functions {yn(x)} that converges to this solution uniformly on I. Example 1: Consider the IVP y0 =3y2/3,y(2) = 0 Then f(x,y)=3y2/3 and @f @y =2y1/3,sof(x,y) is continuous when y = 0 but @f @y is not. Hence the hypothesis of Picard’s Theorem does not hold. Neither does the conclusion; the WebIf y+ y1 =1 then the value of y 3 is A 2 B 1 C -1 D 0 Easy Solution Verified by Toppr Correct option is C) Given y+ y1=1 .................. (1) Multiply by y both the sides Then y 2+1=y ⇒y 2=y−1 ............. (2) multiply by y 2 with (1) Then y 3+y=y 2 ⇒y 3=y 2−y ............ (3) comparing (2) and (3) Then y 3=y−1−y ⇒y 3=−1 Was this answer helpful? WebIf $x = 1$, then $x^2+xy+y^3 = 1$ gives $y+y^3 = 0 \implies y(y^2+1) = 0 \implies y = 0$. In the first step, we get $$2+y' = 0 \implies y' = -2$$ Now we sub. $x = 1, y = 0, y' = -2$ in … darwin\u0027s ideas