Repeat Exercise 27 for the forces KX = (2, – 3) and KY = (- 2, 3).

Q: f(x,y)= eaxy fxx = eaxy ·(24) ²₂ fxy = e² fyy = eaxy. (2x)² fyx = -2xy. 24 axy, fx = e fy= e 2xy.…

A:

Q: (d) 12 2

A: Given, A=132-1

Q: Find u•v

A: As   u = -12,2         v = 1,6   So,  u→=-12 i+2 j        v→= i+6 j

Q: tnx Find if u=5 y=

A: The expression of dydxfor the following function needs to be evaluated : y = ln(x)x5

Q: Assume that u- v = 8, ||u|| = 1, and ||v|| = 10. What is the value of lu- (9u – 8v)?

A: To get the values, first simplify the expression in such a way that you can use the given values.…

Q: Translate Q (0, -8) using (x, y) → (x - 3, y+ 2). Your answer: O Q'(-2, 5) Q' (3, - 10) O Q'(-3, -6)…

A: The transformation is, (x,y) ------> (x-3, y+2) So, (0, -8) ------> (0-3, -8+2)

Q: дz State the multi-variable chain rule for du z=z(t, s), t=t(u, v, w), s=s(u, v, w) if

A:

Q: Əw Əv 2 y Find when u = 0, v = 0 if w=x + + ₁ x= x=5u-5v+1, y = 6u + 6v - 6.

A: To find the indicated partial derivative

Q: Why do we get cov(y0, y0) instead of cov(yt, yt)

A: Given MA(3) process

Q: If u = (u, uz) and v = (V1, V2), how do you find u + v?

A: Given: u=(u1,u2)v=(v1,v2)

Q: Cakulate (3i+2j+k) • (i + zj - k

A:

Q: az Find for z = e2z-y.

A: Given : z = e2x - y

Q: center: (9, –8), vertic@s: (8, –8), (10, –8)

A: