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The two digit number whose ten’s digit is ‘t’ and units’s digit is ‘u’ is 10t + u.

From the question,

Two digit number whose ten’s digit is ‘t’

Two digit number whose unit’s digit is ‘u’

Then, the number = 10 × t + 1 × u

= 10t + u

t + 4, +t – 4, 4 +, t/4, 4/t, -4 – t, 4 + t

(a) 11 added to 2m

2m + 11

(b) 11 subtracted from 2m

2m – 11

(c) 5 times y to which 3 is added

5y + 3

(d) 5 times y from which 3 is subtracted

5y – 3

(e) y is multiplied by -8

-8y

(f) y is multiplied by -8 and then 5 is added to the result

-8y + 5

(g) y is multiplied by 5 and the result is subtracted from 16

16 – 5y

(h) y is multiplied by -5 and the result is added to 16

-5y + 16

(a) 7 added to p

P + 7

(b) 7 subtracted from p

P – 7

(c) p multiplied by 7

7p

(d) p divided by 7

P

(e) 7 subtracted from -m

-m – 7

(f) -p multiplied by 5

– 5p

(g) -p divided by 5

-p/5

(h) p multiplied by -5.

-5p

Let the number be x. 

Thrice of the number is 3x. 

13 subtracted from it is the expression 3x – 13.

Let the number be x. 

Thrice of the number is 3x. 

13 subtracted from it is the expression 3x – 13.

(i) The curve cuts the x-axis at two points. ∴ The equation has 2 zeros. 

(ii) Since the curve cuts the x-axis at 3 different points. The number of zeros of the given curve is three. 

(iii) Since the curve doesn’t cut the x axis. The number of zeros of the given curve is zero. 

(iv) The curve cut the x-axis at one point. ∴ The equation has one zero. 

(v) The curve cut the x axis at one point. ∴ The equation has one zero.

(7x + 2) (3x³ + 2x – x² + 8) = 7x(3x³ + 2x – x² + 8) + 2x 

(3x³ + 2x – x² + 8) = 21x⁴ + 14x² – 7x³ + 56x + 6x³ + 4x – 2x² + 16 = 21x⁴ – x³ + 12x⁴ + 60x + 16

(4) B, A, C, D 

Degree of (A), (B), (C) & (D) are respectively be 5, 4, 8, 12

(3) \(\frac{2}{3}\)

2 – 3x = 0 

-3x = – 2 

x = \(\frac{2}{3}\)

Answer is (4) factor

Let P(x) = 2x³ – 5x² – 28x + 15 

By factor theorem, (x – 5) is a factor of P(x), if P(5) = 0 

P(5) = 2(5)³ – 5(5)² – 28(5) + 15 

= 2 x 125 – 5 x 25 – 140 + 15 

= 250 – 125 – 140 + 15 = 265 – 265 = 0 

∴ (x – 5) is a factor of 2x³ – 5x² – 28x + 15

Let p(x) = x³ + 9x² – x – 105 

By factor theorem, x – 3 is a factor of p(x), if p(3) = 0 

p(3) = 3³ + 9(3)³ – 3 – 105

= 27 + 81 – 3 – 105

 = 108 – 108 

p(3) = 0

To find the zero of x - 3:

Put x - 3 = 0 

We get x = 3

Therefore, x – 3 is a factor of x³ + 9x² – x – 105.

Let p(x) = x³ – 9x² + 26x + k 

By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0 

p(3) = 0 

3³ – 9(3)² + 26(3) + k = 0 

27 – 81 + 78 + k = 0 

k = -24

To find the zero of x - 3:

Put x - 3 = 0 

We get x = 3

(i) 5x – 6 = 0

5x = 6

∴ x = \(\frac{6}{5}\)

(ii) x + 3 = 0

∴ x = -3

(iii) 10x + 9 = 0

10x = -9

∴ x = \(\frac{-9}{10}\)

(iv) 9x – 4 = 0 

9x = 4 

∴ x = \(\frac{4}{9}\)

positive

Since a & b are positive integers,

The solution to the equation ax = b is x = – b/a is also positive.

Let cost of one book be ‘x’

∴ Given that 200 – 7 × x = 60

∴ 200 – 7x = 60

Sum of 3 consecutive integers is 78

Let 1st integer be ‘x’

∴ x + (x + 1) + (x + 2) = 78

∴ x + x + 1 + x + 2 = 78

3x + 3 = 78

Yes, we can get. Consider the below line or equation

x + y = 5

here, when x = 1, y = 4

when x = 2, y = 2

x = 3, y = 2

x = 4, y = 1

Hence, we get multiple solutions for the same linear equation.

Convert to linear equations:

Given that on subtracting 8 from product of 5 and a, we get 32

5 × x – 8 = 32

5x – 8 = 32

Manisha’s uncle is five times of Manisha’s age. Her aunt is 4 years younger than her uncle.

The correct option is (C)  x – 1 = 0.

(C) 2x + 1 = 6 

(B) unknown quantities

In algebra, letters may stand for unknown quantities.

(C) 2x + 1 = 6

Consider the equation, 2x + 1 = 6

Transforming 1 from left hand side to right hand side it becomes -1.

2x = 6 – 1

2x = 5

X = 5/2

(D) x + 13 = 27

Let us assume the number be ‘x’,

Then, adding 13 to the number = x + 13

Therefore, x + 13 = 27

According to the question, sum of zeroes = 3

 Product  of zeroes. = -2 \ 5 

The required quadratic polynomial

= x²- (Sum of zeroes)x + Product of

=x² - 3x - 2 /5

= 1/ 5(5x²- 15x-2)

.'. required quadratic polynomial is 1 /5(5x² - 15x -2)

x times of 3 = 3x

And smallest natural number = 1.

So, 3x added to 1 = 3x + 1.

The required equations are 3y + 4 = 10 and 2x – 3 = 1, i.e., for both equations 2 is the solution.

Area of a rectangle = length x breadth

= (3x + 2) x (3x – 2) = (3x)² – 2² = [9x² – 4] Sq. units 

If x = 20, Area = 9 x 20² – 4 = 9 x 400 – 4 

= 3600 – 4 = 3596 Sq. units

Let P(x) = x² – 2x² – 8 

By using factor theorem,(x + 2) is a factor of P(x), if P (-2) = 0 

P(-2) = (-2)² – 2(-2) – 8 = 4 + 4 – 8 = 0 

And also (x - 4) is a factor of P(x), if P(4) = 0

p(4) = 4² – 2(4) – 8 = 16 – 8 – 8 = 0 

∴ (x + 2), (x – 4) are the sides of a rectangle whose area is x² – 2x – 8.

x²¹⁰⁸ + 2018 is divided by x – 1 

Let g(x) = x – 1 = 0 

x = 1 

p(x) = x²⁰¹⁸ + 2018 

p(1)= 1²⁰¹⁸ + 2018 

= 1 + 2018 = 2019

Let g(x) = x – 2 = 0 

x = 2 

Since p(x) is exactly divisible by (x – 2) 

p(2) = 2(2³) – k(2²) + 3(2)+ 10 

= 16 – 4k + 6 + 10

= 32 – 4k = 0 

= -k = -32 

k = 32/4 = 8.