In Problems 5–14 solve the given linear system.
11.
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A First Course in Differential Equations with Modeling Applications (MindTap Course List)
- _1. The quadratic function y = xi+ 2x-1 is expressed in vertex form y = a(x-h)2+k? a. y=(x+1): + 1 b. y=(x+1): - 2 %3D c. y=(x+1)² + 2 d. y=(x+1): - 1 _2. What is the vertex point of quadratic function y 2x + 4x + 1? b. (1, 1) a. (1, -1) c. (-1, -1) d. (-1, 1) _3. Which of the following is the vertex point of quadratic function y = -2 (x2 -1)? a. (-1,0) b. (1,0) c. (-1, -2) d. (1,-2) 4. Which of the following is the y-intercept of a quadratic function y = (x +1)2 + 3 a.1 b. 2 c.3 d. 4 5. Which of the following table of values represents a quadratic function? a. -2 -1 1 2 3 C. -2 -1 1 2 3 Y 1 2 3 4 5 _Ly -2 -1 1 2 3 b. -3 -2 -1 1 2 d. -2 -1 1 2 3 9. 4 1 1 1 4 -1 -2 -1 7arrow_forwardIn Problems 26–28, find the value of each determinant. |2 1 28. 5 0 2 6 1 4 0 -3 3 4 26. 1 27. -1 2 6 3 4 1 3arrow_forward9. Deternine if S = (7 - 4x + 4x²,6 + 2x − 3x²,20 − 6x + 5x²) is linearly independent or dependentarrow_forward
- Section 2.2 2.1. Solve the following difference equations: (a) Yk+1+Yk = 2+ k, (b) Yk+1 – 2Yk k3, (c) Yk+1 – 3 (d) Yk+1 – Yk = 1/k(k+ 1), (e) Yk+1+ Yk = 1/k(k+ 1), (f) (k + 2)yk+1 – (k+1)yk = 5+ 2* – k2, (g) Yk+1+ Yk = k +2 · 3k, (h) Yk+1 Yk 0, Yk = ke*, (i) Yk+1 Bak? Yk (j) Yk+1 ayk = cos(bk), (k) Yk+1 + Yk = (-1)k, (1) - * = k. Yk+1 k+1arrow_forward12)-7x+/y%3D-19 -2x+3y =-19arrow_forwardIn Problems 15–17, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value. 15. f(x) = 3x² – 6x + 4 16. f(x) = -x² + &x – 4 17. f(x) = -2x² + 4arrow_forward
- 2. LOQ Graphically solve each of the following. (a) 3x + y = 6 and x - y = 2 (b) x + 4y = -8 and 3x + 4y = 0 (c) 5x = 3y and y = -5 (d) 2x + 6y = 8 and x = -2 (e) y = 3x2 and y = 3 (f) y = -2x and x = 4 (g) x = -2 and 3x + 4y = 12 (h) y = -2 and 5x + 3y = 15arrow_forward18. Find the quadratic function from the table of values below. -5 -4 -3 -2 -1 1 -6 -1 2 3 -1 -6 a) y = 2x2 + 4x – 1 b) y = -x2 + 4x – 1 c) y = -x2 – 4x – 1 d) у %3D — 2х2 — 4х - 1 19. What happen to T when h is doubled in the equation T= 4h? b) T is halved a) T is doubled c) T'is tripled d) T becomes zero 20. The number of days needed in repairing a house varies inversely as the number of men working. It takes 15 days for 2 men to repair the house. How many men are needed to complete the job in 6 days? a) 5 men b) 6 men c) 15 men d) 20 menarrow_forwardQuestion. Solve: x" -x + 5y² = + 4" - 4y - 2x² = -2 with x (0) = y(0) = x²(0) = y₁ (0) = 0 Let & [ X (²) } = U (6) & & Gy (+) 2 = V(c) Evaluate U(2) Evaluate (²) Evaluate x(1) Evaluate y(1)arrow_forward
- 3. Determine whether x = T (0 3 0 9 2) is a solution to 3x2 + 2x5 max X1,2,3,4,5 ER subject to x1 + x2x5 x2 + x3 + x5 -2x2 + x4+x5 X1, X2, X3, X4, X5 = = 1 5 5 0 (3)arrow_forwardProblem 16 (#2.3.34).Let f(x) = ax +b, and g(x) = cx +d. Find a condition on the constants a, b, c, d such that f◦g=g◦f. Proof. By definition, f◦g(x) = a(cx +d) + b=acx +ad +b, and g◦f(x) = c(ax +b) + d=acx +bc +d. Setting the two equal, we see acx +ad +b=acx +bc +d ad +b=bc +d (a−1)d=(c−1)b That last step was merely added for aesthetic reasons.arrow_forwardSection 2.2 2.1. Solve the following difference equations: (a) Yk+1+ Yk = 2+ k, (b) Yk+1 – 2yk = k³, (с) ук+1 "Yk = 0, (d) Yk+1 – Yk = 1/k(k+1), (e) Yk+1+ Yk = 1/k(k+1), (f) (k+2)yk+1 – (k + 1)yk = 5 + 2k – k², (g) Yk+1+ Yk = k + 2 · 3k, (h) Yk+1 – Yk = ke“, Yk = Bak*, = cos (bk), (k) Yk+1 + Yk = (-1)*, Yk – k. ,2k (i) Ук+1 (j) Yk+1 – aYk (1) Yk+1 k+1arrow_forward
- Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning