1 -1 is s2 - 16

Q: 14. ) -10 < 4t- 2 s 6

A:

Q: 12-10-1+4

A: Given:

Q: Маax Z-3X,+2X,

A: We will solve this question by graphical method.

Q: :) (;) -2 1 1 х, х(0) — 3 x' = -5 4

A:

Q: 2) 12p'-21p+28p- 49

A: see 2nd step

Q: 40 Συaluate 2k-5 k=1

A:

Q: -4 -2 3 -2 -2 -2 -7 2 -6 Ax2 6 1 12x2 1

A: Complete the given calculations to verify that λ1, λ2, λ3 are the eigen values of A and x1, x2, x3…

Q: [-24 20 -41 -2 1 9. -6 1

A:

Q: 16 - (9 - 4) ÷ 5 =

A:

Q: Graph the functions

A:

Q: -15-3=

A: Given,           -15-3

Q: -12+(4• 3) (-) I 2-14 - 3)ENTER

A:

Q: Find I 25-4 (3-2)(S+)(S-3

A: First, perform the partial fraction decomposition of . Then, use the formula,   to yield the…

Q: Evaluate: t - 1 2 (a) det 3 t - 2

A: Given: Matrix t-123t-2 To determine: Determinant of the given matrix.

Q: -5 8 -7 = A فأن قيمة A12 تكون 1 س أذا كانت 5 -4 -8

A:

Q: -1 1 -1 = -1 -21 2.

A:

Q: (:' } - 8) zp - Vz(z – 2)

A: We use the difference of cube formula for numerator

Q: 2-1 4s - 4 s² - 16/

A:

Q: 3S+9 6- f1 s2-25-3

A:

Q: 2. 9. y =- 4.-72

A: This is a problem of equilibrium and stability analysis.

Q: 13. ) -5 s 3x-8 < 6

A: 13.) given −5 ≤ 3x−8 &lt;6⇒ −5 +8 ≤ 3x &lt; 6+8⇒ 3 ≤ 3x &lt; 14 ⇒1 ≤ x &lt; 14/3

Q: Find y' (-1) if y = In (a² + 5)° (2 – 4æ)®

A: I have stated the used formulas in the solution

Q: 4. y=-글고+3 402-10

A:

Q: 11г - 22 xp- (2г — 1)(х + 5)

A:

Q: 2. z4+5²

A:

Q: 24- 6(-x-3)

A: Solve for x,

Q: Find I} 25-4 3-2)(SHH)(S-3

A:

Q: S 3p + 2q = -1 (5p-3q -27 5. By substitution:

A:

Q: 16 y 18 E F16 10 -10 -5 1 -18 -16

A:

Q: a²²6-1_a-16-2 51. 4a-2-4b-2

A:

Q: 5. +y=ye-² (ve-²) ³ Ans: - e* = ce³x

A: Let us solve the given Ordinary Differential Equation in the next steps.

Q: 4. Evaluate L-1 (s-3)²+8,

A:

Q: 1. x-11<3

A:

Q: -18,-15-12 Calculate T2

A: we have to find out

Q: 7 -18 -2 -3 A = 3 P = -8 -1 -1

A:

Q: 7.) -3- %3D 10 2'

A: .

Q: 2) 12p'-21p' + 28p-49

A:

Q: -1 2 -3 1 2 1 C. -1 2 -3 1 1 6 -1

A:

Q: 2. 3. 13k +5-t -Ik -Ik 12K=-12 K= -/] 5. 25h +40 =-15h-80

A: Solve 3 and 5

Q: -4 -3 -2 -1 -5-4 5 4 3- 2+ 1 -Y -2+ -3+ -4+ -5+ 2 3 4 5 X

A:

Q: 20 z -1

A: Standard form of quadratic equation  ax2+bx+c=0

Q: 10 - v 16 -7 +5+5•4 3

A:

Q: -4a - 8 = -3a + 7

A: The given equation is -4a-8=-3a+7

Q: (11x + 21) 12)- II 11

A: z = 60°  x = 9°

Q: x2= -24 多 (フ-9:)-(-6 ナ3

A: We will solve the given problem by using properties of complex numbers.

Q: 2. (x- ブメッテルーョ fx 3.

A:

Q: 1- 4an-2, ao =

A:

Q: х ах 13. S- (6х-x2-5)2 3

A:

Q: (3/-6) (-11V-10) 11V - 10

A: To calculate the value of the provided expression which is shown below,

Q: (4) S . Hint: = | (-). (x²2. -4)² (교2-4)2 %3D x+2

A: