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(x - 6)(x - 6)By component wise multiplication

= x(x - 6) - 6(x - 6) 

(from above we can see that, x.x = x², x.(-6) 

= - 6x, -6.x = - 6x, and -6.-6 = +36) 

= x² - 6x - 6x + 36 

= x² - 12x + 36 

Note: Multiplication of signs is given by-

(+) × (+) = + 

(+) × (-) = - 

(-)×(+)= - 

(-) × (-) = +

Given ((x/y)-((y/x))²

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

((x/y)-((y/x))²= (x/y)²-2 (x/y)( y/x)+(y/x)²

((x/y)-((y/x))²= x²/y²-2+y²/x²

Given (x+3) (x-3)

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

Applying the formula we get

(x+3) (x-3) =x²-3²

(x+3) (x-3) = x²-9

A) x² + (a + b) x + ab 

A) (a + b) (a – b)

Correct option is (A) (a + b) (a – b)

\(96\times104\) = (100 - 4) \(\times\) (100 + 4)

Let a = 100, b = 4

Then \(96\times104\) = (a - b) (a + b)

\(\therefore\) Formula used in product of \(96\times104\) is (a + b) (a – b).

Correct option is  B) identity

(0.8x – 0.5y) (1.5x – 3y)

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= 0.8x, b= 0.5y, c= 1.5x, d= 3y

Now,

= 0.8x × (1.5x – 3y) – 0.5y × (1.5x – 3y)

= [(0.8x × 1.5x) + (0.8x × -3y)] – [(0.5y × 1.5x) + (0.5y × -3y)]

= [1.2x² – 2.4xy – 0.75yx + 1.5y²]

= [1.2x² – 3.15xy + 1.5y²]

After solving the equation, 

we get,

(x + 5)(x – 3) = x(x – 3) + 5(x – 3) 

= x² – 3x + 5x – 15 

= x² + 2x – 15

We know that,

(x + 4)(x + 4) = (x + 4)²

From formula, (a + b)² = a² + b² + 2ab

(x + 4)² = x² + 4² + 2 × x × 4

= x² + 8x + 16

After solving the equations, 

we get, 

(2x + 3)(3x – 1) = 2x(3x – 1) + 3(3x – 1) 

= 6x² – 2x + 9x – 3 

= 6x² + 7x – 3

Distance = speed × time 

d = s × t

Simple interest  = (Principal x Time x Rate of interest) / 100

∴ I  = \(\frac{PRT}{100}\)

By using horizontal method, 

We have; 

= (3p² + q²) × (2p² – 3q²) 

= 3p²(2p² – 3q²) + q²(2p² – 3q²) 

= 6p⁴ – 9p²q² + 2p²q² – 3q⁴ 

= 6p⁴ – 7p²q² – 3q⁴

By using horizontal method,

We have;

= (x³ – y³) × (x² + y²) 

= x³(x² + y²) – y³(x² + y²) 

= x⁵ + x³y² – x²y³ - y⁵

By using horizontal method,

We have;

(9x² – x + 15) × (x² – x – 1)

= 9x²(x² – x – 1) – x (x² – x – 1) +15(x² – x – 1)

= 9x⁴ – 9x³ – 9x² – x³ + x² + x + 15x² – 15x – 15

Putting equal power terms together, we get,

= 9x⁴ – 9x³ – x³ – 9x² + x² + 15x² – 15x + x – 15

= 9x⁴ – 10x³ + 7x² – 14x – 15

By using horizontal method, 

We have; 

(2x²+ 3x – 7) ×(3x² – 5x + 4) 

= 2x²(3x² – 5x + 4) + 3x(3x² – 5x + 4) – 7(3x² – 5x + 4) 

= 6x⁴ – 10x³ + 8x² + 9x³ – 15x² + 12x– 21x² + 35x – 28 

Now, putting equal power terms together, we get, 

= 6x⁴ – 10x³ + 9x³ + 8x² – 15x²– 21x² + 35x + 12x– 28 

= 6x⁴ – x³ – 28x² + 47x – 28

By using horizontal method, 

We have; 

= (x⁴ + y⁴) × (x² – y²) 

= x⁴(x² – y²) + y⁴(x² – y²) 

= x⁶ – x⁴y² + x²y⁴ – y⁶

By using horizontal method,

We have;

= (4x + 9) × (x - 6) 

= 4x(x – 6) + 9(x - 6) 

= 4x² – 24x + 9x – 54 

= 4x² – 15x – 54

By using horizontal method, 

We have; 

= (x² – 3x + 7) × (2x + 3) 

= 2x(x² – 3x + 7) + 3(x² – 3x + 7) 

= 2x³ - 6x² + 14x + 3x² – 9x + 21 

By arranging the expression in the form of descending powers of x, 

We get; 

= 2x³ – 6x² + 3x² + 14x – 9x + 21 

= 2x³ – 3x² + 5x + 21

By using horizontal method, 

We have; 

= (x² – xy + y²) × (x + y) 

= x(x² – xy + y²) + y(x² – xy + y²) 

= x³ - x²y + xy² + x²y – xy² + y³ 

By arranging the expression in the form of descending powers of x, 

We get; 

= (x³ + y³)

By using horizontal method, 

We have; 

= (2x + 5) × (4x - 3) 

= 2x (4x – 3) + 5 (4x – 3) 

= 8x² - 6x + 20x – 15 

= 8x² + 14x - 15

Given (x⁴+y⁴) × (x² –y²)

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,

(x⁴+y⁴) × (x² –y²)

⇒ x⁴ (x² –y²) + y⁴(x² –y²)

⇒ x⁶ – x⁴y² + x²y⁴ – y⁶

By using horizontal method, 

We have; 

= (x³ – 2x² + 5) × (4x – 1) 

= 4x(x³ – 2x² + 5) – 1(x³ – 2x² + 5) 

= 4x⁴ – 8x³ + 20x – 1x³ + 2x² – 5 

By arranging the expression in the form of descending powers of x, 

We get; 

= 4x⁴ – 8x³ – x³ + 2x² + 20x – 5 

= 4x⁴ – 9x³ + 2x² + 20x – 5

By using horizontal method, 

We have;

= (x² + xy + y²) × (x - y) 

= x(x² + xy + y²) - y(x² + xy + y²) 

= x³ + x²y + xy² - x²y – xy² - y³ 

By arranging the expression in the form of descending powers of x, 

We get; 

= (x³ - y³)

xy (x³ – y³)

= xy × x³ – xy × y³

= x⁴y – xy⁴

Correct option is  A) – 15xy

(-13/5)ab²c × (7/3)a²bc²

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(-13/5) × (7/3)] × (a × a²) × (b² × b) × (c × c²)

= [(-13×7)/ (5×3)] × (a¹⁺²) × (b²⁺¹) × (c¹⁺²) … [∵ a^m × aⁿ = a^m+n]

= [-91/ 15] × (a³) × (b³) × c³

= [-91/15]a³b³c³

 (2/3)x²y × (3/5)xy²

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(2/3) × (3/5)] × (x² × x) × (y × y²)

= [(2×3)/ (3×5)] × (x²⁺¹) × (y¹⁺²) … [∵ a^m × aⁿ = a^m+n]

= [(2×1)/ (1×5)] × (x³) × (y³)

= [2/5]x³y³

C) 6p, – 7p, \(\frac{5}{2}\)p 

Correct option is  D) 9x²

Correct option is (D) 9x²

A + B = \((5x^2+3xy+2y^2)\) \(+(-2y^2–3xy+4x^2)\)

\(=(5+4)x^2+3xy-3xy+2y^2-2y^2\)

\(=9x^2.\)

Correct option is  A) 5

Correct option is (A) 5

Degree = Largest exponent in the polynomial

Here, largest exponent is \(x^5.\)

So, degree = 5.

Correct option is  B) \(\frac{-3}{4}\) xy 

Correct option is  C) 6

Correct option is (C) 6

Degree of \(4xy^2z^3\) = 1+2+3 = 6.

Given -72x²y²z by -12xyz

⇒-72x²y²z / (-12xyz)

On dividing monomial by a monomial we have divide same variables of each

Expressions

On simplifying we get

⇒6xy

(i) By dividing 24x²y³ by 3xy

We get;

\(=\frac{24\text{x}^2y^3}{3\text{x}y}\)

= 8xy²

(ii) By dividing 36xyz² by by – 9xz

We get;

\(=\frac{36\text{x}yz^2}{-9\text{x}z}\)

= - 4yz

(iii) By dividing – 72x²y²z by – 12xyz

We get;

= \(\frac{-72\text{x}^2y^2z}{-12\text{}\text{x}yz}\)

= 6xy

(iv) By dividing – 56mnp² by 7mnp

We get;

= \(\frac{-56mnp^2}{7mnp}\)

= - 8p

Given 8x²y²-6xy²+10 x²y³ by 2xy

⇒8x²y²-6xy²+10 x²y³/ (2xy)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

⇒4xy-3y+5xy²

i. x + 7 = 4 

∴ x + 7 – 7 = 4 – 7 ….(Subtracting 7 from both sides) 

∴ x + 0 = -3 

∴ x = -3

ii. 4p = 12

∴ 4p/4 = 12/4 

….(Dividing both sides by 4)

∴ p = 3

iii. m – 5 = 4 

∴ m – 5 + 5 = 4 + 5 

…. (Adding 5 to both sides)

∴ m + 0 = 9 

∴ m = 9

iv. t/3 = 6

∴ t/3 × 3 = 6 × 3 

…. (Multiplying both sides by 3)

∴ t = 18

(a) Monomial and polynomial in y

Consider the given equation, x⁴ – xy + 2y² and –x⁴ + xy + 2y²

Sum of two expressions = (x⁴ – xy + 2y²) + (–x⁴ + xy + 2y²)

= x⁴ – xy + 2y² – x⁴ + xy + 2y²

= (x⁴ – x⁴)+ (-xy + xy) + (2y² + 2y²)

= 0 + 0 + 4y²

= 4y²

(c) 5 times of y = 5y

Now, subtraction of 5 times of y from x is written as x - 5y.

(c) x

An irreducible factor is a factor which cannot be expressed further as a product of factors. Such a factorisation is called an irreducible factorisation.

24x²y² = 2 × 2 × 2 × 3 × x × x × y × y

Therefore an irreducible factor is x.

False.

Sum of x² + x and y + y²

= (x² + x) + (y + y²)

= x² + y² + x + y

Correct option is  B) Degree

Correct option is C) ax² + bx + c

Correct option is  A) 2

i. 16xy × 18xy 

= 16 × 18 × xy × xy 

= 288x²y² 

ii. 23xy² × 4yz² 

= 23 × 4 × xy² × yz² 

= 92xy³z² 

iii. (12a + 17b) × 4c 

= 12a × 4c + 17b × 4c 

= 48ac + 68bc

iv. (4x + 5y) × (9x + 7y) 

= 4x × (9x + 7y) + 5y × (9x + 7y) 

= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y) 

= 36x² + 28xy + 45xy + 35y² 

= 36x² + 73xy + 35y²

(i) Given a² + b² -2a² + c² + 4a

The like terms in the given algebraic expressions are a² and −2a².

(ii) Given 3x + 4xy − 2yz + 52zy

The like terms in the given algebraic expressions are -2yz and 5/2zy.

(iii) Given abc + ab²c + 2acb² + 3c²ab + b²ac − 2a²bc + 3cab²

The like terms in the given algebraic expressions are ab²c, 2acb², b²ac and 3cab².

C) The length of your classroom is fixed

As per the given information, taxi service charges Rs. 8 per km and fixed charge of ? 50.
If taxi is hired for x km.

Then, algebraic expression for the situation = 8 x x + 50 = 8x + 50 Hence, the required expression is 8x + 50.

Given, money paid to shiv = Rs. 50 per h.

∴ Money paid last week =Rs. 50 x 7 = Rs. 350

So, money paid this week = Rs. 50 x x = Rs. 50x

Total money paid to shiv =Rs. (350 + 50x) =Rs. 50(x + 7)

Correct option is D) All the above

(i) y²x + y

Terms which contains x - y²

co-efficient of x - y²

(ii) 13y² - 8yx

Terms which contains x- (x - 8yx)

co-efficient of x - (-8y)

(iii) x + y + 2

Terms which contains x - x

co-efficient of x -1

(iv) 5 + z + zx

Terms which contains x - x

co-efficient of x - z

(v) 1 + x + xy

Terms which contains x - x and xy

co-efficient of x - 1 y

(vi) 12xy² + 25 - 12xy²

Terms which contains x

co-efficient of x - 12y²

(vii) 7x + xy²

7x Terms which contains x

co-efficient of x - 7

Terms which contains x - xy²

y² co-efficient of