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Correct option is  D) not defined

Given (4x + 9) × (x – 6)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(4x + 9) × (x – 6)

⇒4x (x – 6) + 9 (x – 6)

⇒4x² – 24x + 9x – 54

⇒ 4x² – 15x – 54

Given (69)²

But we can write 69 as 70-1

And also we know that (a – b)² = a²-2ab+b²

By applying the above identity we get

(69)² = (70-1)²= 70²-2(70) (1) +1²

(70-1)²=4900-140+1=4761

(82)² – (18)² 

By using (a – b)(a + b) 

= a² – b² = (82 – 18)(82 + 18) 

= (64)(100) 

= 6400

We can write following problem such as, 

(197 × 203) = (200 – 3)(200 + 3) 

From the formula, (a +b)(a – b) = a² – b² 

We get, 

(200 – 3)(200 + 3) = 200² – 3² = 40000 – 9 = 39991.

Given (2x + 5) × (4x – 3)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(2x + 5) × (4x – 3)

⇒2x (4x – 3) + 5 (4x – 3)

⇒8x² – 6x + 20x – 15

⇒ 8x² + 14x – 15

Correct option is  C) a + 2b

Correct option is (C) a + 2b

2a + b – (a – b) = 2a+b-a+b

= a + 2b

Given that (x²+7) (x²+7)

But we can write the given expression as

(x²+7) (x²+7)= (x²+7)²

But we have (a + b)²=a²+2ab+b²

On applying above identity in the given expression we get,

(x²+7)²= (x²)²+2 ((x²) (7) + (7)²

(x²+7)²= x⁴+14x² + 49

i) (a – b)² = a² – 2ab + b²

a = 3, b = 1

⇒ (3 – 1)² = (3)² – 2 × 3 × 1 + 1²

⇒ (2)² = 9 – 6 + 1

⇒ 4 = 4

∴ (i) is an identity,

ii) (a + b) (a – b) = a² – b²

a = 2, b = 1

⇒ (2 + 1) (2 – 1) = (2)² – (1)²

⇒ 3 × 1 = 4 – 1

⇒ 3 = 3

∴ (ii) is an identity.

(k + 3m) (3m – k) 

= k(3m – k) + 3m(3m – k)

= k × 3m – k × k + 3m × 3m – 3m × k

= 3km – k² + 9m² – 3km

= 9m² – k²

Given that ((5/6) a²+2) ((5/6) a²+2)

But we can write the given expression as

((5/6) a²+2) ((5/6) a²+2)= ((5/6) a²+2)²

But we have (a + b)²=a²+2ab+b²

On applying above identity in the given expression we get,

((5/6) a²+2)²= ((5/6) x)²+2 ((5/6)) (2) + (2)²

((5/6) a²+2)²= (25/36) x²+ (10/3) x + 4y²

Correct option is  A) 7x + 9

Correct option is (A) 7x + 9

\(2x^2+5x-1+8x+x^2+7–6x+3–3x^2\)

\(=(2+1-3)x^2+(5+8-6)x+(-1+7+3)\)

\(=0x^2+7x+9\)

= 7x + 9

Given (5x² + (3/4)y²) (5x²-(3/4)y²)

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

(5x² + (3/4) y²) (5x²-(3/4) y²)= (5x²)²-((3/4) y²)²

(5x² + (3/4) y²) (5x²-(3/4) y²)= 25x⁴-(9/16) y⁴

Given (x²y – yz²)^-2

According to the identity (a – b)²=a²-2ab+b² we have to expand the given expression,

(x²y – yz²)^-2= (x²y)²-2 (x²y)( yz²)+( yz²)²

(x²y – yz²)^-2= x⁴y²-2x²y²z²+y²z⁴

Given (3y – 8) × (5y – 1)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3y – 8) × (5y – 1)

⇒3y (5y – 1) – 8 (5y – 1)

⇒15y² – 3y – 40y + 8

⇒15y² – 43y + 8

Given (x – 4) (x – 4)

But we can write the given expression as (x – 4) (x -4) = (x – 4)²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

(x – 4)²= x²-2 x (4) + 4²

(x – 4)²= x²-8 x + 16

 Given (7x + 2y) × (x + 4y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(7x + 2y) × (x + 4y)

⇒7x (x + 4y) + 2y (x + 4y)

⇒7x² + 28xy + 2yx + 8y²

⇒ 7x² + 39xy + 8y²

By using horizontal method, 

We have; 

= (3y - 8) × (5y - 1) 

= 3y(5y – 1) – 8(5y – 1) 

= 15y² – 3y – 40y + 8 

= 15y² – 43y +8

By using horizontal method, 

We have; 

= (7x + 2y) × (x + 4y) 

= 7x(x +4y) + 2y(x + 4y) 

= 7x² + 28xy + 2xy + 8y² 

= 7x² + 30xy + 8y²

Length of the rectangle = (8x + 5) cm 

Breadth of the rectangle = (5x + 3) cm 

∴ Area of the rectangle = length × breadth 

= (8x + 5) × (5x + 3) 

= 8x × (5x + 3) + 5 × (5x + 3) 

= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3) 

= 40x² + 24x + 25x + 15 

= 40x² + 49x + 15 

∴ The area of the rectangle is (40x² + 49x + 15) sq. cm.

Given (2x-3y) (2x-3y)

But we can write the given expression as (2x – 3y) (2x -3y) = (2x – 3y)²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

(2x – 3y)²= 4x²-2 (2x) (3y) + 9y²

(2x – 3y)²= 4x²-12 xy + 9y²

Correct answer is

(A) -34 x⁵y⁴z³

Correct answer is

(D) -2(7x + 12y)

Hints:

(3x – 11y) – (17x + 13y) 

= 3x – 11y – 17x – 13y 

= – 14x – 24y 

= – 2 × 7x – 2 × 12y 

= – 2(7x + 12y)

-5a(7a – 2b)

= -5a × 7a – (-5a) × 2b

= -5 × 7 × a × a + 5 × 2 × a × b

= -35a² + 10ab

Product of given Equation is=     -35a² + 10ab

\(\frac{6}{5}x(x^3+y^3)\)

= \(\frac{6}{5}x\times x^3+\frac{6}{5}x\times y^3\)

= \(\frac{6}{5}x^4+\frac{6}{5}xy^3\)

Given that ((3/4) x – (5/6) y) ((3/4) x + (5/6) y)

But we can write the given expression as

((3/4) x – (5/6) y) ((3/4) x + (5/6) y)= ((3/4) x – (5/6) y)²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

((3/4) x – (5/6) y)²= ((3/4) x)²-2 ((3/4) x) ((5/6) y) + ((5/6) y)²

((3/4) x – (5/6) y)²= (9/16) x²– (15/12) x + (25/36)y²

By using horizontal method,

We have;

= (2x² – 5y²) × (x² + 3y²) 

= 2x²(x² + 3y²) – 5y²(x² + 3y²) 

= 2x⁴ + 6x²y² – 5x²y² – 15y⁴ 

= 2x⁴ + x²y² – 15y⁴

(i) -5y² from y²

= y² – (-5y²)

= y² + 5y² = 6y²

(ii) 6xy from – 12xy

= -12xy – (6xy)

= -12xy – 6xy = -18xy

(iii) (a – b) from (a + b)

= a + b – (a – b) 

= a + b - a + b

= 2b

(iv) a(b – 5) from b(5 – a)

= ab – 5a from 5b – ab

= (5b – ab) – (ab – 5a)

= 5b – ab – ab + 5a

= 5a – 5b – 2ab

(v) -m² + 5mn from 4m² – 3mn + 8

= (4m² – 3mn + 8) – (-m² + 5mn)

= 4m² – 3mn + 8 + m² – 5mn

= 4m² + m² – 3mn – 5mn + 8

= 5m² – 8mn + 8

(vi) -x² + 10x – 5 from 5x – 10

= (5x – 10) – (-x² + 10x – 5)

= 5x – 10 + x² – 10x + 5

= 5x – 10x + x² – 10 + 5

= x² – 5x – 5

(vii) 5a² – 7 ab + 5b² from 3ab – 2a² – 2b²

= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)

= 3ab – 2a² – 2b² – 5a² + 7ab – 5b

= -2a² – 5a² – 2b² – 5b² + 7ab + 3ab

= -7a² – 7b² + 10ab

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq

= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)

= 5p² + 3q² – pq² – 4pq + 5q² + 3p²

= 5p² + 3p² + 3q² + 5q² – 4pq – pq

= 8p² + 8q² – 5pq

Correct option is B) 22 xy

Given that (x – (3/x)) (x – (3/x))

But we can write the given expression as

(x – (3/x)) (x – (3/x))= (x – (3/x))²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

(x – (3/x))²= (x)²-2 (x) (3/x) + (3/x)²

(x – (3/x))²= x²– 6+ (9/x²)

By using horizontal method, 

We have; 

= (3m – 4n) × (2m – 3n) 

= 3m(2m – 3n) – 4n(2m – 3n) 

= 6m² – 9mn – 8mn + 12n² 

= 6m² – 17mn + 12n²

By using horizontal method, 

We have; 

= (9x + 5y) × (4x + 3y) 

= 9x(4x + 3y) + 5y(4x + 3y) 

= 36x² + 27xy + 20xy + 15y² 

= 36x² + 47xy + 15y²

(5y – 1) (3y – 8)

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c – d) = a × (c – d) – b × (c – d) = (a × c – a × d) – (b × c – b × d)

= ac – ad – bc + bd

Let,

a= 5y, b= 1, c= 3y, d= 8

Now,

= 5y × (3y – 8) -1 × (3y – 8)

= [(5y × 3y) + (5y × -8)] – [(1 × 3y) + (1 × -8)]

= [15y² – 40y – 3y + 8]

= [15y² – 43y + 8]

Given that ((1/3) x² – 9) ((1/3) x² – 9)

But we can write the given expression as

((1/3) x² – 9) ((1/3) x² – 9)= ((1/3) x² – 9)) ²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

((1/3) x² – 9))²= ((1/3) x)²-2 ((1/3) x) (9) + (9)²

((1/3) x² – 9))²= ((1/9) x⁴)- 6x +81

(7x + 2y) (x + 4y)

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d)

= ac + ad + bc + bd

Let,

a= 7x, b= 2y, c= x, d= 4y

Now,

= 7x × (x + 4y) +2y × (x + 4y)

= [(7x × x) + (7x × 4y)] + [(2y × x) + (2y × 4y)]

= [7x² + 28xy + 2yx + 8y²]

= [7x² + 30xy + 8y²]

The perimeter of the rectangle ABCD

= \(\overline{AB} + \overline{BC} + \overline{CD} + \overline{DA} \) 

or 

= 2 (l + b) 

= 2 (3x + 2x) 

= 2 × 5x 

= 10x units

The perimeter of a rectangle = 2 (length + breadth)

= 2 (6x + y + 3x – 2y)

= 2 (9x – y)

∴ P = (18x – 2y) units

Given that ((1/2) y² – (1/3) y) ((1/2) y² – (1/3) y)

But we can write the given expression as

((1/2) y² – (1/3) y) ((1/2) y² – (1/3) y) = ((1/2) y² – (1/3) y) ²

But we have (a – b)²=a²-2ab+b²

On applying above identity in the given expression we get,

((1/2) y² – (1/3) y) ²= ((1/4) y²-2 ((1/2) y) (1/3) + (1/3) y²

((1/2) y² – (1/3) y) ²= ((1/4) y²– y (1/3) + (1/9) y²

Given x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5) 

= x + y + 1 + x + 2 + y + 3 + x + 4 + y + 5

= (x + x + x) + (y + y + y) + (1 + 2 + 3 + 4 + 5) 

= 3x + 3y + 15

Given (8a+3b)²

According to the identity (a + b)²=a²+2ab+b² we have to expand the given expression,

(8a+3b)²= (8a)²+2 (8a)(3b)+(3b)²

(8a+3b)²= 64a²+48ab+9b²

Given (5x+11)²

According to the identity (a + b)²=a²+2ab+b² we have to expand the given expression,

(5x+11)²= (5x)²+2 (5x)(11)+(11)²

(5x+11)²= 25x²+110x+121

Correct option is  C) ≡

Given (7x+2y)²

According to the identity (a + b)²=a²+2ab+b² we have to expand the given expression,

(7x+2y)²= (7x)²+2 (7x)(2y)+(2y)²

(7x+2y)²= 49x²+28xy+4y²

Given ((a/2) + (2/a))²

According to the identity (a + b)²=a²+2ab+b² we have to expand the given expression,

((a/2) + (2/a))²= (a/2)²+2 (a/2) (2/a)+ (2/a)²

((a/2) + (2/a))²= a²/4+2+4/a²

Given (9x-10)²

According to the identity (a – b)²=a²-2ab+b² we have to expand the given expression,

(9x-10)²= (9x)²-2 (9x)(10)+(10)²

(9x-10)²= 81x²-180x+100

Given ((3x/4) + (2y/9))²

According to the identity (a + b)²=a²+2ab+b² we have to expand the given expression,

((3x/4) + (2y/9))²= (3x/4)²+2 (3x/4) (2y/9)+ (2y/9)²

((3x/4) + (2y/9))²= 9x²/16+ (1/3) xy + (4y ²/81)

Correct option is  C) a² – b² 

We know that,

From formula, (a + b)(a – b) = a² – b²

(2x + 5)(2x – 5) = (2x)² – (5)²

= 4x² – 25

D) a² – 2ab + b²