4n+3 IIi II i = k=1

Q: 2a + 56+ c -a – 36 – 2c) 3a + 7b +c -4a – 96 - c, 2. T(а + bx + сх*).

A: The solution is given by using rank nullity theorem and null space as follows :

Q: 3

A:

Q: 16. Σ 3k-1 k=1

A:

Q: ||v|| = 4, 0 = 3.4°

A: Given query is to find component of v.

Q: 2+(29-xy)

A: We first get rid of the bracket.  So,  2+(29-xy)=2+29-xy

Q: 2-5s 5. (s-6)(s²+16)

A:

Q: (4 -1 1 (b).| 3 7 - 2 X2 1 3 1 II

A:

Q: = V3 + 4" + 7" and b, = Vn2 + 3n – Vn2 – 5n.

A:

Q: x²e 6. S (х2+1)2

A: Given : ∫x3ex2x2+12dx

Q: 2x²12-59 P₁ = (x, y) 2 y = 4x P₂=(x, y)

A: See the details solution in below

Q: z²+1 5. S ₂²³(x+5) I

A: Let's solve given integral.

Q: 3n n-9 54 46 А B Зх-1 4 34 5x-2

A:

Q: Example 4: Sohue: (2+j3)2²+ 4z +4-j6=0

A:

Q: 4sin^(2)(x)-3=0

A:

Q: 6. Solv gx=12- 3 (b) 18, smi 2*+ 2

A:

Q: 1) x² – y² = 25 , y" %3D

A:

Q: (n-2n-25n2 -80n+65)+(n-7)

A:

Q: 2v-9=7(v+9)

A: Given:

Q: 2+ 2r -+34 dr. -12 1. Find

A: ∫2x3+2x2-x+34x4-x2-12dx=∫2x3+2x2-x+34x4-x2-12dx=∫x-4x2+3-1x+2+2x-2dx=∫x-4x2+3dx-∫1x+2dx+2∫1x-2dx Now…

Q: 2. ä – 4.i = t2 3. ä + i – 2x = 3e¬t 4. ä + à – 2x = 3et

A:

Q: a1 = -31, d = -7, n = 5 O-231 0-230 O1-225 -240

A:

Q: V 48° 2 38° Z.

A: The given circle is: As we know the sum of angles of a triangle is 180°.

Q: (4m-3n -7 2m +3n = 19 6. By elimination:

A:

Q: (k) Yk+1+Yk = (-1)*,

A: We have to do proper summation here.

Q: 2n+2 Σ (3k + 2n) = = k=0 = || ⠀ first step 2nd step 3rd step final answer

A: We know that 1+2+3+4+......+n is equal to n(n+1)/2

Q: 24h5i4j2

A: 36h⁸ i⁴ j⁶ ÷ 24h⁵ i⁴ j² = (1.5)h³ i⁰ j⁴ = 3h³j⁴/2

Q: 1 + 3-1 V3+1 = q +3 V3 %3D

A: P=5, q=2

Q: an = 2an-1 + 8a,-2 + 81n for n > 2 with initial conditions ao = :-23, aj = -59. An =

A:

Q: 5s' -7s+17 b- 1". (s-1)(s' +4)

A: First of all we will do partial fraction, then we will apply kappas inverse

Q: 4a² • 3a5 = (4 + 3)a² + 5 = 7a 7 %3D %3D

A: The solution are as follows

Q: Find mZR. 4/43 8-/43 R mZR

A:

Q: | grk(1 – 2)n=k/(") z*(1 – 2)n-k | 2*(1 – : 1/2 1 2n/2 how did we get step 3 from step 2 ? explain…

A: First consider the second term of step 2 , which is ∑k=0nnkxk1-xn-k12 This term is in the form of…

Q: у 7. y' +²√y = 0, y(1) = 0 - x

A:

Q: Solve: I 5x + 3 I = I 12 - 4x I

A: Given that the equation I 5x + 3 I = I 12 - 4x I To solve the given equation

Q: V3, 3 2/3; + 8; i - k 4 1 83B k

A:

Q: 2. 2 y /0/0 II II 2/3

A:

Q: 2s – 4 (s² + s)(s² + 1) 2

A:

Q: (b) х + 2х — 3%3D 0

A: We can solve the given problem using concept of roots of polynomial of degree 3. According to the…

Q: Simplify:  (3+ 2n2 +4n4)+ (n3 − 7n2 − 4n4)

A:

Q: 7n3 + 2n An Зп3 — 4 3n3 - 4

A:

Q: (b) у" — 4у'+ Зу %3D Зе2х + 4е* + 2e3х

A: Given differential equation: y''-4y'+3y=3e2x+4ex+2e3xTo find: General solution of the differential…

Q: (ii)""+2n-1 n' +3

A: We have to test the series is convergent or divergent

Q: k c2(k) = -(-1)* = -1½[1+ (–1)*] + c2, i=0

A: Given,  c2(k) = -∑i=0k(-1)i = -121+-1k+c2 Expand the summation  -∑i=0k(-1)i…

Q: x²+3 20. x-i

A: x2+3x-i=x2+3x-i×x+ix+i           =x2+3x+x2+3ix2-i2           =x3+3x+ix2+i3x2--1

Q: 5) V-20+9r =r %3D

A: Given problem :- Solve  √(-20+9r) =r

Q: 25. y = V2r + 3 %3D

A: Since you have asked multiple question, we will solve the first question for you. If youwant any…

Q: 1. V5 I (V +1- Vr2 1) VI - 1

A: topic - definite integrals