Rea -10 -5 10 ax -4-2 V2 i -5- 10

Q: 2 3 (25-x2) 2

A: Given:The integral is,

Q: 18 9s 52-4s+4 s2-4s+4

A: Consider the given expression. 9ss2-4s+4-18s2-4s+4   The objective of the problem is to simplify the…

Q: b15-スx2+ x 4x t.

A:

Q: 2+(29-xy)

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

Q: 1 -3s s2-

A:

Q: 8(a+9) =

A: Given,           8(a + 9)

Q: 8n+5 II (1+i) = k=1

A: To find ∏k=18n+51+i:

Q: Expand -5x (2 - 3y)

A: Expand -5x (2 - 3y)

Q: Evaluate 10s? – 5 - L-1 2 (s² + 1)(s² + 3)²

A:

Q: 4x+8 x²+7x+12 x2-x-6 x²-9

A:

Q: -3p-3/4 - 30p-7/4

A: We have to find the value of p as nothing else is mentioned in the question

Q: 1-0 1-V

A: The given function for integration is,

Q: 2(3+d-1)

A: Solve the given expression  given expression is 2(3+d-1)

Q: 5+2-5+p 13) 3. 3) 28

A: Given: The equation 53p+23=5+p2p, To determine: The value of p,

Q: 3x+5 / 3x-1 = 5x+3 / x+1

A: Solve the given equation

Q: Find the Partial derivative of f with respect to y for * 2. y² f (x, y) = x + y

A:

Q: 2 7t 3/2 dt

A: we will use substitution

Q: 4 1 x-1 4 | 45. +3 x-1

A:

Q: 5x-2x-19 9) Evaluate (x+3(x-1)

A:

Q: x +10x +16 x+3 x? +2x-15 6 -6x-16 x² -9 x +4x -5 x+8 x² -64 5.

A:

Q: X+2 1-く

A: Here, we consider the two cases. 1) x+2>0. 2)x+2<0

Q: (8,-1)+(2,1)

A:

Q: +6 12+ +10+ 1- 1. 4- 3- X0X 1-

A: We have a 5×5 kenken puzzle. We will put values from 1 to 5 in the cells satisfying the given…

Q: 14 2x+ 3(x-2) 3(x-2)

A:

Q: 6(2x + 5)  = 10(3 + x) +10 then  x =

A: With the use of disturbutive law we can easily solve.

Q: 9-4(10-x)=6x+1

A: The given expression is 9 – 4(10 – x) = 6x + 1.

Q: Expand (x²n + 2x"y" + y2n)(x2n – 2x"y" + y2n)

A: To evaluate: x2n+2xnyn+y2nx2n-2xnyn+y2n We know that xm×xn = xm+n

Q: Expand (2x+y)5

A: