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over the compact region. −1 ≤ x ≤ 1, 0 ≤ y ≤ 2. Solution: We look for the critical points in the interior: ∇f = (2x − 2+2y,4y − 2+ ... Let second-degree equation be, ax 2 + by 2 + 2hxy + 2gx + 2fy + c = 0. Discriminant = Δ = Case: 1 If discriminant (Δ) of this equation doesn't May 29, 2012 ... The point (0,0) is on the curve. Since the equation is quadratic, for each slope t, the line y=tx must intersect the curve at a second point. If ... 2x + 4y = 12. Solve for x & y. 5x + 4y = 16. Noting that the ... 0 = (x + 2)(x − 3). Knowing that y = 6, and x = −2,3, we can express ... Substitute (y−2)2−4 ( y - 2 ) 2 - 4 for y2−4y y 2 - 4 y in the equation x2+y2−2x−4y=4 x 2 + y 2 - 2 x - 4 y = 4 . 1 Answer. Alan P. Dec 5, 2015. (x−1)2+(y−(−2))2=(√6) ... ... 2x 8 y. 5 x. 5. ( 1, 0). (0, 1). (1, 0) y x2 1 y. 5 x. 5. ( 2, 0). (0, 4). (2, 0) ... (a) T1x2 = 12 - x. 5. + 2 x2 + 4. 3. (b) 5x 0 … x … 126. (c) 3.09 hr (d) 3.55 ... x²+3x+2. 2x-4. 20. x²-4. 8 a+3. 244 ба. 21. 00. 2(x+3)(x-3). 4(x+1). (x-2)² eng ... #4 y= y≤ x+2. -늘X+늘. Dec 3, 2021 ... x^2+y^2 +2x - 4y - 8 = 0 Here are 2 methods: 1️⃣ Compare quadratic x²+y²+2fx+2gy+c = 0 & centre (-f,-g) 2f = 2 ⟹ f =1 also 2g = -4 ⟹ g = -2 so ... 4 9 XY X 4 Y 9 49315 9 49 2 Z4 2 Z[ \4 4 1 9 ]1 Z19 Z4 5 4 54 9 5 31 4 31 2 Z ^1 5 2219 5 Y 5 31 19 Y Z42 45Y 9 994 1 9 Z45 Z4 Z ^1 [ Z454 4 1^24Y 54 49 2 9919 ...
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