Explain five major factors which contributed to the making of our constitution. -Maths 9th

Explain five major factors which contributed to the making of our constitution. -Maths 9th
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Answer :

The white minority and the black majority agreed to draw up a common constitution for South Africa in the following ways.The white regime changed its repressive policies and repealed its discriminatory laws.Ban on political parties and restrictions on media were lifted.The black leaders appealed to fellow blacks to forgive the whites for the atrocities committed by them.There were long sessions of debates and discussions and they searched a solution to the existing problems.It was agreed that everyone should be a part of the solution irrespective of the differences and whatever they had done or represented in the past
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Description : 1. Which substance found in blood which helps in cloting? 2. How can a major portion of the Constitution be amended? 3. What is the product of the time period of a wave and its frequency? 4. ... ? 20. Which region of northern India was not included in the empire of Ala-ud-din Khalji ?

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find the mean of first five multiples of 10 -Maths 9th

Description : find the mean of first five multiples of 10 -Maths 9th

Last Answer : This is the correct one...

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

find the mean of first five multiples of 10 -Maths 9th

Description : find the mean of first five multiples of 10 -Maths 9th

Last Answer : This is the correct one...

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

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The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

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The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Last Answer : (c) Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 ⇒ x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Last Answer : Excluded number is

Find the arithmetic mean of first-five natural numbers. -Maths 9th

Description : Find the arithmetic mean of first-five natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5/5 = 15/5 = 3

Find the value of 7plus root five ÷by seven minus root five minus seven minus root five ÷by seven plus root five equals to a+7÷11root five b -Maths 9th

Description : Find the value of 7plus root five ÷by seven minus root five minus seven minus root five ÷by seven plus root five equals to a+7÷11root five b -Maths 9th

Last Answer : This answer was deleted by our moderators...

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Description : Five years hence , the age of Ram will be 10 more than the two thirds of Ravi’s age . Assume the present ages of Ram and Ravi as x and y respectively . Express the statement in the form of a linear equation in two variables. -Maths 9th

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Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

Description : Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

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The figure given besides shows five squares inside one another by joining the midpoints of the outer square. -Maths 9th

Description : The figure given besides shows five squares inside one another by joining the midpoints of the outer square. -Maths 9th

Last Answer : (c) \(rac{31}{16}a^2.\)Side of the square I = a ∴ Area of square I = a2= \(\sqrt{rac{a^2}{16}+rac{a^2}{16}}\) = \(\sqrt{rac{2a^2}{16}}\) = \(rac{a}{2\sqrt2}\)∴ Sum of the areas of the five squares= a2 + \( ... = a2 \(\bigg[1+ rac{1}{2} + rac{1}{4}+rac{1}{8}+rac{1}{16}\bigg]\) = \(rac{31}{16}a^2.\)

A can hit a target three times in five shots, B two times in five shots and C three times in four shots. They fire a volley. -Maths 9th

Description : A can hit a target three times in five shots, B two times in five shots and C three times in four shots. They fire a volley. -Maths 9th

Last Answer : P(A) = \(rac{3}{5}\), P(B) = \(rac{2}{5}\), P(C) = \(rac{3}{4}\). In order that two shots may hit the target the following three mutually exclusive events are possible,(i) A and B hit the target and not C. (ii) B ... {100}\) + \(rac{12}{100}\) + \(rac{27}{100}\) = \(rac{45}{100}\) = \(rac{9}{20}.\)

A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Description : A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Last Answer : Total number of ways in which 5 tickets can be drawn = n(S) = 30C5. The tickets are arranged in the form T1, T2, T3 (= 20), T4, T5 Where T1, T2 ∈{1, 2, 3, , 19} and T4, T5 ∈{21, 22, , 30 ... {10 imes9}{2}\) x \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Description : A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Last Answer : Without repetition, a five -digit number can be formed using the five digits in 5! ways (5 4 3 2 1) Out of these 5! numbers, 4! numbers will be starting with digit 0. (0 (fixed) 4 3 2 1) ∴ Total ... + 6 + 6 + 4 + 4 + 4 = 30∴ Required probability = \(rac{30}{96}\) = \(rac{5}{16}.\)

Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time? -Maths 9th

Description : In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time? -Maths 9th

Last Answer : ∴Required number of ways = 7C4 = 7!

There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. -Maths 9th

Description : There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. -Maths 9th

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Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Description : Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Last Answer : According to the question, we have 5 balls to be placed in 3 boxes where no box remains emptyHence, we can have the following kinds of distribution firstly, where the distribution will be (3,1,1) that is, one box gets three ... go in second box is = 4 C 2 . Total no. of ways =90 Total:60+90=150.

Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Description : Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Last Answer : this is the ans hope its clear

Assess the contribution made by Constituent Assembly to form the Indian Constitution. [HOTS] The following were the contributions made by the Constituent Assembly to form the Indian Constitution. -Maths 9th

Description : Assess the contribution made by Constituent Assembly to form the Indian Constitution. [HOTS] The following were the contributions made by the Constituent Assembly to form the Indian Constitution. -Maths 9th

Last Answer : answer:

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.