Total number of balls = 12 Total number of black balls = x P (getting a black ball) = x/12 If 6 more black balls are put in the box, then Total number of balls = 12 + 6 = 18 Total number of black balls = x + 6 P (getting a black ball now) = (x + 6)/18 According to the condition given in the questionRead more

Total number of balls = 12
Total number of black balls = x
P (getting a black ball) = x/12
If 6 more black balls are put in the box, then
Total number of balls = 12 + 6 = 18
Total number of black balls = x + 6
P (getting a black ball now) = (x + 6)/18
According to the condition given in the question,
2(x/12) = (x + 6)/18
⇒ 3x = x + 6
⇒ 2x = 6
⇒ x = 3

Let the number of blue balls be x. Number of red balls = 5 Total number of balls = x + 5 P (getting a red all) = 5/(5 + x) P (getting a blue ball) = x/(5 + x) Given that, 2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x - 50 = 0 ⇒ x² - 10x + 5x - 50 = 0 ⇒ x(x - 10) + 5(x - 10) = 0 ⇒ (x - 10)(x + 5)Read more

Let the number of blue balls be x.
Number of red balls = 5
Total number of balls = x + 5
P (getting a red all) = 5/(5 + x)
P (getting a blue ball) = x/(5 + x)
Given that,
2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x – 50 = 0
⇒ x² – 10x + 5x – 50 = 0 ⇒ x(x – 10) + 5(x – 10) = 0
⇒ (x – 10)(x + 5) = 0
⇒ Either x – 10 = 0 or x + 5 = 0
⇒ x – 10 or x = -5
However, the number of balls cannot be negative.
Hence, number of blue balls = 10

QR is diameter of circle. Therefore, ∠RPQ = 90° [Angle in semicircle is right angle] In APQR, by Pythagoras theorem, RP² + PQ² = RQ² (7)² + (24)² = RQ² ⇒ RQ² = 576 + 49 = 625 ⇒ RQ = √625 = 25 Therefore, the radius of circle = RQ/2 = 25/2 cm Area of shaded region = Area of semicircle - Area of APQR =Read more

QR is diameter of circle.
Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]
In APQR, by Pythagoras theorem,
RP² + PQ² = RQ²
(7)² + (24)² = RQ²
⇒ RQ² = 576 + 49 = 625
⇒ RQ = √625 = 25
Therefore, the radius of circle = RQ/2 = 25/2 cm
Area of shaded region = Area of semicircle – Area of APQR
= 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12
= 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = (-1)/4 = (-b)/a αβ = 1/4 = c/a On comparing, a = 4, b = 1 and c = 1 Hence, the required quadratic polynomial is 4x² + x + 1. For Video Solution See Here 👀

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = (-1)/4 = (-b)/a
αβ = 1/4 = c/a
On comparing,
a = 4, b = 1 and c = 1
Hence, the required quadratic polynomial is 4x² + x + 1.

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 4 = 4/1 = (-b)/a αβ = 1 = 1/1 = c/a On comparing, a = 1, b= - 4 and c = 1 Hence, the required quadratic polynomial is x² - 4x + 1. See Here 😃✌

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = 4 = 4/1 = (-b)/a
αβ = 1 = 1/1 = c/a
On comparing,
a = 1, b= – 4 and c = 1
Hence, the required quadratic polynomial is x² – 4x + 1.

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = √2 = (3√2)/3 = (-b)/a αβ = 1/3 = c/a On comparing, a = 3, b =-3/√2 and c = 1 Hence, the required quadratic polynomial is 3x² - 3/√2 + 1. Video Explanation 😃 Understanding polynomials is essential for further study in matRead more

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = √2 = (3√2)/3 = (-b)/a
αβ = 1/3 = c/a
On comparing,
a = 3, b =-3/√2 and c = 1
Hence, the required quadratic polynomial is 3x² – 3/√2 + 1.

Video Explanation 😃

Understanding polynomials is essential for further study in mathematics, including topics such as algebra and calculus. Studying polynomials helps students develop important mathematical skills, such as algebraic manipulation, problem-solving, and critical thinking.
See less

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 1 = 1/1 = (-b)/a αβ = 1 = 1/1 c/a On comparing, a = 1, b = -1 and c = 1 Hence, the required quadratic polynomial is x² - x + 1. Explanation 👇

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = 1 = 1/1 = (-b)/a
αβ = 1 = 1/1 c/a
On comparing,
a = 1, b = -1 and c = 1
Hence, the required quadratic polynomial is x² – x + 1.

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 1/4 = (-b)/a αβ = -1 = (-4)/4 = c/a On comparing, a = 4, b = -1 and c = -4 Hence, the required quadratic polynomial is 4x² - x - 4. See here 👇

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = 1/4 = (-b)/a
αβ = -1 = (-4)/4 = c/a
On comparing,
a = 4, b = -1 and c = -4
Hence, the required quadratic polynomial is 4x² – x – 4.

3x² - x - 4 = 3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1) The value of 3x² - x - 4 is zero if 3x - 4 0 or x + 1 0. ⇒ x = 4/3 or x = -1. Therefore, the zeroes of 3x² - x - 4 are and -1. Now, Sum of zeroes = 4/3 +(-1) = (4 - 3)/3 = 1/3 = -(-1)/3 = -(Coficient of x)/(Cofficient of x²) PRead more

3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
The value of 3x² – x – 4 is zero if 3x – 4 0 or x + 1 0.
⇒ x = 4/3 or x = -1.
Therefore, the zeroes of 3x² – x – 4 are and -1.
Now, Sum of zeroes = 4/3 +(-1) = (4 – 3)/3 = 1/3 = -(-1)/3 = -(Coficient of x)/(Cofficient of x²)
Product of zeroes = 4/3 × (-1) = -4/3 = (-4)/3 = (Coefficient of term)/(Cofficient of x²)

## A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Total number of balls = 12 Total number of black balls = x P (getting a black ball) = x/12 If 6 more black balls are put in the box, then Total number of balls = 12 + 6 = 18 Total number of black balls = x + 6 P (getting a black ball now) = (x + 6)/18 According to the condition given in the questionRead more

Total number of balls = 12

Total number of black balls = x

P (getting a black ball) = x/12

If 6 more black balls are put in the box, then

Total number of balls = 12 + 6 = 18

Total number of black balls = x + 6

P (getting a black ball now) = (x + 6)/18

According to the condition given in the question,

2(x/12) = (x + 6)/18

⇒ 3x = x + 6

⇒ 2x = 6

⇒ x = 3

See here for video explanation🙌

See less## A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Let the number of blue balls be x. Number of red balls = 5 Total number of balls = x + 5 P (getting a red all) = 5/(5 + x) P (getting a blue ball) = x/(5 + x) Given that, 2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x - 50 = 0 ⇒ x² - 10x + 5x - 50 = 0 ⇒ x(x - 10) + 5(x - 10) = 0 ⇒ (x - 10)(x + 5)Read more

Let the number of blue balls be x.

See lessNumber of red balls = 5

Total number of balls = x + 5

P (getting a red all) = 5/(5 + x)

P (getting a blue ball) = x/(5 + x)

Given that,

2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x – 50 = 0

⇒ x² – 10x + 5x – 50 = 0 ⇒ x(x – 10) + 5(x – 10) = 0

⇒ (x – 10)(x + 5) = 0

⇒ Either x – 10 = 0 or x + 5 = 0

⇒ x – 10 or x = -5

However, the number of balls cannot be negative.

Hence, number of blue balls = 10

## Find the area of the shaded region in Figure.

QR is diameter of circle. Therefore, ∠RPQ = 90° [Angle in semicircle is right angle] In APQR, by Pythagoras theorem, RP² + PQ² = RQ² (7)² + (24)² = RQ² ⇒ RQ² = 576 + 49 = 625 ⇒ RQ = √625 = 25 Therefore, the radius of circle = RQ/2 = 25/2 cm Area of shaded region = Area of semicircle - Area of APQR =Read more

QR is diameter of circle.

Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]

In APQR, by Pythagoras theorem,

RP² + PQ² = RQ²

(7)² + (24)² = RQ²

⇒ RQ² = 576 + 49 = 625

⇒ RQ = √625 = 25

Therefore, the radius of circle = RQ/2 = 25/2 cm

Area of shaded region = Area of semicircle – Area of APQR

= 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12

= 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²

Here is the video explanation ✌😇

See less## Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. -1/4,1/4

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = (-1)/4 = (-b)/a αβ = 1/4 = c/a On comparing, a = 4, b = 1 and c = 1 Hence, the required quadratic polynomial is 4x² + x + 1. For Video Solution See Here 👀

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have

α + β = (-1)/4 = (-b)/a

αβ = 1/4 = c/a

On comparing,

a = 4, b = 1 and c = 1

Hence, the required quadratic polynomial is 4x² + x + 1.

For Video Solution See Here 👀

See less## Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 4,1

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 4 = 4/1 = (-b)/a αβ = 1 = 1/1 = c/a On comparing, a = 1, b= - 4 and c = 1 Hence, the required quadratic polynomial is x² - 4x + 1. See Here 😃✌

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have

α + β = 4 = 4/1 = (-b)/a

αβ = 1 = 1/1 = c/a

On comparing,

a = 1, b= – 4 and c = 1

Hence, the required quadratic polynomial is x² – 4x + 1.

See Here 😃✌

See less## Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: p(x) = x ³ – 3x² + 5x – 3, g(x) = x² – 2

Here is the Video Solution 😃

Here is the Video Solution 😃

See less## Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. √(2 ) ,1/3

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = √2 = (3√2)/3 = (-b)/a αβ = 1/3 = c/a On comparing, a = 3, b =-3/√2 and c = 1 Hence, the required quadratic polynomial is 3x² - 3/√2 + 1. Video Explanation 😃 Understanding polynomials is essential for further study in matRead more

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have

α + β = √2 = (3√2)/3 = (-b)/a

αβ = 1/3 = c/a

On comparing,

a = 3, b =-3/√2 and c = 1

Hence, the required quadratic polynomial is 3x² – 3/√2 + 1.

Video Explanation 😃

Understanding polynomials is essential for further study in mathematics, including topics such as algebra and calculus. Studying polynomials helps students develop important mathematical skills, such as algebraic manipulation, problem-solving, and critical thinking. See less

## Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1,1

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 1 = 1/1 = (-b)/a αβ = 1 = 1/1 c/a On comparing, a = 1, b = -1 and c = 1 Hence, the required quadratic polynomial is x² - x + 1. Explanation 👇

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have

α + β = 1 = 1/1 = (-b)/a

αβ = 1 = 1/1 c/a

On comparing,

a = 1, b = -1 and c = 1

Hence, the required quadratic polynomial is x² – x + 1.

Explanation 👇

See less## Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.1/4, -1

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 1/4 = (-b)/a αβ = -1 = (-4)/4 = c/a On comparing, a = 4, b = -1 and c = -4 Hence, the required quadratic polynomial is 4x² - x - 4. See here 👇

Let α and β are the zeroes of the polynomial ax² + bx + c, then we have

α + β = 1/4 = (-b)/a

αβ = -1 = (-4)/4 = c/a

On comparing,

a = 4, b = -1 and c = -4

Hence, the required quadratic polynomial is 4x² – x – 4.

See here 👇

See less## Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 3x² – x – 4.

3x² - x - 4 = 3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1) The value of 3x² - x - 4 is zero if 3x - 4 0 or x + 1 0. ⇒ x = 4/3 or x = -1. Therefore, the zeroes of 3x² - x - 4 are and -1. Now, Sum of zeroes = 4/3 +(-1) = (4 - 3)/3 = 1/3 = -(-1)/3 = -(Coficient of x)/(Cofficient of x²) PRead more

3x² – x – 4

= 3x² – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4)(x + 1)

The value of 3x² – x – 4 is zero if 3x – 4 0 or x + 1 0.

⇒ x = 4/3 or x = -1.

Therefore, the zeroes of 3x² – x – 4 are and -1.

Now, Sum of zeroes = 4/3 +(-1) = (4 – 3)/3 = 1/3 = -(-1)/3 = -(Coficient of x)/(Cofficient of x²)

Product of zeroes = 4/3 × (-1) = -4/3 = (-4)/3 = (Coefficient of term)/(Cofficient of x²)

See Here 😃

See less