To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C, such that BC = 1 unit. Draw the perpendicular bisector of AC which intersects AC at 0. Taking O as centre and OA as radius, draw a semi-circle which intersects D to the perpendicular at B. Now taking 0 as centre and OD aRead more

To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C, such that BC = 1 unit. Draw the perpendicular bisector of AC which intersects AC at 0. Taking O as centre and OA as radius, draw a semi-circle which intersects D to the perpendicular at B. Now taking 0 as centre and OD as radius, draw an arc, which intersects AC produced at E. Hence, OE = √9.3.

With a scale or tape we get only an approximate rational number as the result of our measurement. That is why n can be approximately represented as a quotient of two rational numbers. As a matter of mathematical truth it is irrational.

With a scale or tape we get only an approximate rational number as the result of our measurement. That is why n can be approximately represented as a quotient of two rational numbers. As a matter of mathematical truth it is irrational.

I studied this solution from NCERT solution for class 10 Maths. We know that, P(not E) = 1 - P(E) P(not E) =1- 0.05 = 0.95 Therefore, the probability of 'not E' is 0.95. Check the step by step solutions in the video here

I studied this solution from NCERT solution for class 10 Maths.
We know that, P(not E) = 1 – P(E)
P(not E) =1- 0.05 = 0.95
Therefore, the probability of ‘not E’ is 0.95.
Check the step by step solutions in the video here

It is not an equally likely event. as it depends on the player's ability and there is no information given about that. If you still want to know about the detailed answer you can see the solution is the video.

It is not an equally likely event. as it depends on the player’s ability and there is no information given about that. If you still want to know about the detailed answer you can see the solution is the video.

## Represent √9.3 on the number line.

To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C, such that BC = 1 unit. Draw the perpendicular bisector of AC which intersects AC at 0. Taking O as centre and OA as radius, draw a semi-circle which intersects D to the perpendicular at B. Now taking 0 as centre and OD aRead more

To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C, such that BC = 1 unit. Draw the perpendicular bisector of AC which intersects AC at 0. Taking O as centre and OA as radius, draw a semi-circle which intersects D to the perpendicular at B. Now taking 0 as centre and OD as radius, draw an arc, which intersects AC produced at E. Hence, OE = √9.3.

See less## Rationalise the denominators of the following: 1/√7.

(i) 1/√7 = 1/√7 × √7/√7 = √7/7

(i) 1/√7 = 1/√7 × √7/√7 = √7/7

See less## Rationalise the denominators of the following: 1/(√7 – 2).

(ii) 1/(√7 - √6) = 1/(√7 - √6) × (√7 + √6)/(√7 + √6) = (√7 + √6)/((√7)² - (√6)²)) = (√7 + √6)/(7 - 6) = √7 + √6

(ii) 1/(√7 – √6) = 1/(√7 – √6) × (√7 + √6)/(√7 + √6) = (√7 + √6)/((√7)² – (√6)²)) = (√7 + √6)/(7 – 6) = √7 + √6

See less## Simplify each of the following expression: (√5 – √2) (√5 + √2).

(iv) (√5 - √2)(√5 + √2) = (√5)² - (√2)² = 5 - 2 = 3 [∵ (a - b)(a + b) = a² - b²]

(iv) (√5 – √2)(√5 + √2) = (√5)² – (√2)² = 5 – 2 = 3 [∵ (a – b)(a + b) = a² – b²]

See less## Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π= c/d. This seems to contradict the fact that πis irrational. How will you resolve this contradiction?

With a scale or tape we get only an approximate rational number as the result of our measurement. That is why n can be approximately represented as a quotient of two rational numbers. As a matter of mathematical truth it is irrational.

With a scale or tape we get only an approximate rational number as the result of our measurement. That is why n can be approximately represented as a quotient of two rational numbers. As a matter of mathematical truth it is irrational.

See less## Simplify each of the following expression: (3+√3)(2+√2).

(i) (3 + √3)(2 + √2) = 6 + 3√2 + 2√3 + 2√3 + √6

(i) (3 + √3)(2 + √2) = 6 + 3√2 + 2√3 + 2√3 + √6

See less## Simplify each of the following expression: (3 + √3) (3 – √3)

(ii) (3 + √3)(3 - √3) = 3² - (√3)² [∵ (a + b)(a - b) = a² - b²] = 9 - 3 = 6

(ii) (3 + √3)(3 – √3) = 3² – (√3)² [∵ (a + b)(a – b) = a² – b²] = 9 – 3 = 6

See less## Simplify each of the following expressions: (√5 + √2)²

(iii) (√5 + √2)² = (√5)² + (√2)² + 2 × √5 × √2 = 7 + 2√10 [∵ (a + b)² = a² + b² + 2ab]

(iii) (√5 + √2)² = (√5)² + (√2)² + 2 × √5 × √2 = 7 + 2√10 [∵ (a + b)² = a² + b² + 2ab]

See less## If P(E) = 0.05, what is the probability of ‘not E’?

I studied this solution from NCERT solution for class 10 Maths. We know that, P(not E) = 1 - P(E) P(not E) =1- 0.05 = 0.95 Therefore, the probability of 'not E' is 0.95. Check the step by step solutions in the video here

I studied this solution from NCERT solution for class 10 Maths.

See lessWe know that, P(not E) = 1 – P(E)

P(not E) =1- 0.05 = 0.95

Therefore, the probability of ‘not E’ is 0.95.

Check the step by step solutions in the video here

## A player attempts to shoot a basketball. She/he shoots or misses the shot. Is it equally likely outcomes?

It is not an equally likely event. as it depends on the player's ability and there is no information given about that. If you still want to know about the detailed answer you can see the solution is the video.

It is not an equally likely event. as it depends on the player’s ability and there is no information given about that. If you still want to know about the detailed answer you can see the solution is the video.

See less