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.
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 lessRationalise 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 lessRationalise 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 lessSimplify 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 lessRecall, 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