The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH}} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 - 1/4 = 3/4
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
Number of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}}
P (Hanif will win the game) = 2/8 = 1/4
P (Hanif will lose the game) = 1 – 1/4 = 3/4
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm. Area of semicircle = 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm² Area of square = (side)² = (14)² = 196 cm Area of shaded region = Area of square - Area of two semicircles = 196 - 2 × 77 = 196 - 154 42 cm² See this videoRead more
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm.
Area of semicircle
= 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm²
Area of square
= (side)² = (14)² = 196 cm
Area of shaded region = Area of square – Area of two semicircles
= 196 – 2 × 77 = 196 – 154 42 cm²
Radius of smaller circle = 7 cm Radius of larger circle = 14 cm Area of shaded region = Area of sector OAFCO - Area of sector OBEDO = 40°/360° × π(14)² - 40°/360° × π(7)² = 1/9 × 22/7 × 14 × 14 - 1/9 × 22/7 × 7 × 7 = 616/9 - 154/9 = 462/9 = 154/3 cm² Here is the video explanation of the above questiRead more
Radius of smaller circle = 7 cm
Radius of larger circle = 14 cm
Area of shaded region
= Area of sector OAFCO – Area of sector OBEDO
= 40°/360° × π(14)² – 40°/360° × π(7)²
= 1/9 × 22/7 × 14 × 14 – 1/9 × 22/7 × 7 × 7
= 616/9 – 154/9 = 462/9 = 154/3 cm²
Here is the video explanation of the above question😀👇
Radius of each quadrant = 1 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm² Area of circle = πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm² Area of square = (Side)² = (4)² = 16 cm² Area of shaded region = Area of square Area of circle - Area of 4 quadrants 16 - 22/7 -Read more
Radius of each quadrant = 1 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm²
Area of circle
= πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm²
Area of square = (Side)² = (4)² = 16 cm²
Area of shaded region = Area of square Area of circle – Area of 4 quadrants
16 – 22/7 – 4 × 11/14 = 16 – 22/7 – 22/7 = (112 – 44)/7 = 68/7 cm²
LHS = sin((B+C)/2) = sin ((180°-A)/2) [∵ A +B +C = 180°] = sin (90°- A/2) = cos A/2 = RHS [∵ sin (90° - 0) = cos 0] Here is the explanation video of the above question😎👇
LHS = sin((B+C)/2)
= sin ((180°-A)/2) [∵ A +B +C = 180°]
= sin (90°- A/2)
= cos A/2 = RHS [∵ sin (90° – 0) = cos 0]
Here is the explanation video of the above question😎👇
We know that sin θ = cos (90° - θ) and cos θ = sin (90° - θ) Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° - 75°) = cos 23° + sin 15° Explanation video🙌😀
We know that sin θ = cos (90° – θ) and cos θ = sin (90° – θ)
Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° – 75°)
= cos 23° + sin 15°
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero. See here the Solution 👀
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero.
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² - 4x + 4 has one real zero (2), while the cubic polynomial x³ - 2x² + x - 1 has no real zeros. See here for Solution👇
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² – 4x + 4 has one real zero (2), while the cubic polynomial x³ – 2x² + x – 1 has no real zeros.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH}} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 - 1/4 = 3/4
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
See lessNumber of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}}
P (Hanif will win the game) = 2/8 = 1/4
P (Hanif will lose the game) = 1 – 1/4 = 3/4
Find the area of the shaded region in Figure.
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm. Area of semicircle = 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm² Area of square = (side)² = (14)² = 196 cm Area of shaded region = Area of square - Area of two semicircles = 196 - 2 × 77 = 196 - 154 42 cm² See this videoRead more
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm.
Area of semicircle
= 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm²
Area of square
= (side)² = (14)² = 196 cm
Area of shaded region = Area of square – Area of two semicircles
= 196 – 2 × 77 = 196 – 154 42 cm²
See this video explanation 😎👇
See lessFind the area of the shaded region in Figure.
Radius of smaller circle = 7 cm Radius of larger circle = 14 cm Area of shaded region = Area of sector OAFCO - Area of sector OBEDO = 40°/360° × π(14)² - 40°/360° × π(7)² = 1/9 × 22/7 × 14 × 14 - 1/9 × 22/7 × 7 × 7 = 616/9 - 154/9 = 462/9 = 154/3 cm² Here is the video explanation of the above questiRead more
Radius of smaller circle = 7 cm
Radius of larger circle = 14 cm
Area of shaded region
= Area of sector OAFCO – Area of sector OBEDO
= 40°/360° × π(14)² – 40°/360° × π(7)²
= 1/9 × 22/7 × 14 × 14 – 1/9 × 22/7 × 7 × 7
= 616/9 – 154/9 = 462/9 = 154/3 cm²
Here is the video explanation of the above question😀👇
See lessFrom each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Figure. Find the area of the remaining portion of the square.
Radius of each quadrant = 1 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm² Area of circle = πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm² Area of square = (Side)² = (4)² = 16 cm² Area of shaded region = Area of square Area of circle - Area of 4 quadrants 16 - 22/7 -Read more
Radius of each quadrant = 1 cm
See lessArea of each quadrant
= 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm²
Area of circle
= πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm²
Area of square = (Side)² = (4)² = 16 cm²
Area of shaded region = Area of square Area of circle – Area of 4 quadrants
16 – 22/7 – 4 × 11/14 = 16 – 22/7 – 22/7 = (112 – 44)/7 = 68/7 cm²
If A, B and C are interior angles of a triangle ABC, then show that Sin(B+C/2) = cos A/2.
LHS = sin((B+C)/2) = sin ((180°-A)/2) [∵ A +B +C = 180°] = sin (90°- A/2) = cos A/2 = RHS [∵ sin (90° - 0) = cos 0] Here is the explanation video of the above question😎👇
LHS = sin((B+C)/2)
= sin ((180°-A)/2) [∵ A +B +C = 180°]
= sin (90°- A/2)
= cos A/2 = RHS [∵ sin (90° – 0) = cos 0]
Here is the explanation video of the above question😎👇
See lessExpress sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
We know that sin θ = cos (90° - θ) and cos θ = sin (90° - θ) Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° - 75°) = cos 23° + sin 15° Explanation video🙌😀
We know that sin θ = cos (90° – θ) and cos θ = sin (90° – θ)
Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° – 75°)
= cos 23° + sin 15°
Explanation video🙌😀
See lessIf tan A = cot B, prove that A + B = 90°.
Given that: tan A = cot B ⇒ cot (90° - A) = cot B [∵ cot (90° - θ) = tan θ] ⇒ 90° - A = B ⇒ 90° = A + B Hence, A + B = 90° Video explanation 👇
Given that: tan A = cot B
⇒ cot (90° – A) = cot B [∵ cot (90° – θ) = tan θ]
⇒ 90° – A = B ⇒ 90° = A + B
Hence, A + B = 90°
Video explanation 👇
See lessIf sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Given that: sec 4A = cosec(A - 20°) ⇒ cosec(90° - 4A) = cosec(A - 20°) [∵ cosec(90°- θ) = sec θ] ⇒ 90°-4A = A - 20° ⇒ 90°+20° = 5A ⇒ 5A = 110° A = 22° Hence, A = 22° See this for better understanding😮👇
Given that: sec 4A = cosec(A – 20°)
⇒ cosec(90° – 4A) = cosec(A – 20°) [∵ cosec(90°- θ) = sec θ]
⇒ 90°-4A = A – 20°
⇒ 90°+20° = 5A ⇒ 5A = 110° A = 22°
Hence, A = 22°
See this for better understanding😮👇
See lessVerify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 2x³ + x² – 5x + 2; ½,1, -2
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero. See here the Solution 👀
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero.
See here the Solution 👀
See lessVerify that the numbers given alongside of the cubic polynomial below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: x³ – 4x² + 5x – 2; 2, 1, 1
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² - 4x + 4 has one real zero (2), while the cubic polynomial x³ - 2x² + x - 1 has no real zeros. See here for Solution👇
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² – 4x + 4 has one real zero (2), while the cubic polynomial x³ – 2x² + x – 1 has no real zeros.
See here for Solution👇
See less