√3-1÷√3+1 rationalise (can solve by finding the rationalizing factor of denominator -Maths 9th

√3-1÷√3+1 rationalise (can solve by finding the rationalizing factor of denominator -Maths 9th
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√3-1÷√3+1 rationalise (can solve by finding the rationalizing factor of denominator -Maths 9th

Description : √3-1÷√3+1 rationalise (can solve by finding the rationalizing factor of denominator -Maths 9th

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

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Simplify by rationalizing the denominator : -Maths 9th

Description : Simplify by rationalizing the denominator : -Maths 9th

Last Answer : Rationalizing the denominator

Simplify by rationalizing the denominator : -Maths 9th

Description : Simplify by rationalizing the denominator : -Maths 9th

Last Answer : Rationalizing the denominator

Rationalise the denominator of the following -Maths 9th

Description : Rationalise the denominator of the following -Maths 9th

Last Answer : Denominator of the following

Rationalise the denominator of the following -Maths 9th

Description : Rationalise the denominator of the following -Maths 9th

Last Answer : Denominator of the following

Rationalize the denominator -Maths 9th

Description : Rationalize the denominator -Maths 9th

Last Answer : open the bracket and change signs and multiply it by both denominator as well as numerator.

Rationalize the denominator -Maths 9th

Description : Rationalize the denominator -Maths 9th

Last Answer : open the bracket and change signs and multiply it by both denominator as well as numerator.

Maths | Complex Numbers | If √3 + i = r (cosθ + i sinθ), then find the value of θ in radian measure.

Description : Maths | Complex Numbers | If √3 + i =  r (cosθ + i sinθ), then find the value of θ in radian measure.

Last Answer : Solution :

Class 12 | Maths | Complex Numbers | Express the complex numbers modulus-amplitude form - √3 + i

Description : Class 12 | Maths | Complex Numbers | Express the complex numbers modulus-amplitude form  - √3 + i

Last Answer : Solution : 

What is an example of fractions that you would compare by finding common denominator and an example of fractions you would compare by finding common numerators?

Description : What is an example of fractions that you would compare by finding common denominator and an example of fractions you would compare by finding common numerators?

Last Answer : Need answer

Does anyone actually want to help the poor, or is it all just rationalizing?

Description : Does anyone actually want to help the poor, or is it all just rationalizing?

Last Answer : I think what you're doing is completely awesome. I can say that I do care, and panhandlers definitely don't scare me. Portland has a kind of alarming rate of homelessness for a city ... perspective other than lazy bum! , but I hope that mindset completely disappears with the younger generations.

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : NEED ANSWER

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : The child does not understand, that data has to be arranged in ascending or descending order before finding the median.

Without finding the cubes, factorise: (2r-3s)3 +(3s -5t)3+ (5t-2r)3. -Maths 9th

Description : Without finding the cubes, factorise: (2r-3s)3 +(3s -5t)3+ (5t-2r)3. -Maths 9th

Last Answer : Solution :-

Using identities solve 103^ -Maths 9th

Description : Using identities solve 103^ -Maths 9th

Last Answer : NEED ANSWER

Using identities solve 103^ -Maths 9th

Description : Using identities solve 103^ -Maths 9th

Last Answer : FRIEND YOUR QUESTION IS NOT CLEAR

Solve for x: 5(4x + 3) = 3(x -2) -Maths 9th

Description : Solve for x: 5(4x + 3) = 3(x -2) -Maths 9th

Last Answer : Solution :-

Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Description : Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Last Answer : Solution :-

Solve for x: Sol. 3x/7 + 2/7 + 4(x + 1)/5 = 2/3(2x + 1) -Maths 9th

Description : Solve for x: Sol. 3x/7 + 2/7 + 4(x + 1)/5 = 2/3(2x + 1) -Maths 9th

Last Answer : Solution :-

Solve the equation u-5 =15 and state the axiom that you use here. -Maths 9th

Description : Solve the equation u-5 =15 and state the axiom that you use here. -Maths 9th

Last Answer : Solution :-

Solve for x if a > 0 and 2 logx a + logax a + 3 loga2x a = 0 -Maths 9th

Description : Solve for x if a > 0 and 2 logx a + logax a + 3 loga2x a = 0 -Maths 9th

Last Answer : (d) \(a^{-rac{4}{3}}\)Since logaxa = \(rac{1}{log_aax}\) = \(rac{1}{log_aa+log_ax}\) = \(rac{1}{1+log_ax}\) andloga2x a = \(rac{1}{log_aa^2x}\) = \(rac{1}{log_aa^2+log_{ax}x}\) = \(rac{1}{2log_aa+log_{ax}x}\)= \( ... 2}}\)When t = \(-rac{4}{3}\), logax = \(-rac{4}{3}\) ⇒ \(x\) = \(a^{-rac{4}{3}}\)

Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Description : Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them -Maths 9th

Description : A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them -Maths 9th

Last Answer : Let E be the event that A solve the problem and F the event that B solves the problem.Then P(E) = \(rac{90}{100}\) = \(rac{9}{10}\), P(F) =\(rac{70}{100}\) = \(rac{7}{10}\), P(\(\bar{E}\) ... probability that at least one of them will solve a problem = 1 - \(rac{3}{100}\) = \(rac{97}{100}\) = 0.97.

Solve the following linear inequations: -Maths 9th

Description : Solve the following linear inequations: -Maths 9th

Last Answer : answer:

Solve the following pairs of inequations and also graph the solution set -Maths 9th

Description : Solve the following pairs of inequations and also graph the solution set -Maths 9th

Last Answer : answer:

Solve x^2 – 5x + 4 > 0. -Maths 9th

Description : Solve x^2 – 5x + 4 > 0. -Maths 9th

Last Answer : answer:

Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th

Description : Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th

Last Answer : Case 1. |x-1|-3>0 |x+2|–53, |x+2|

Solve : 4^(1 + x) + 4^(1 – x) = 10 for x. -Maths 9th

Description : Solve : 4^(1 + x) + 4^(1 – x) = 10 for x. -Maths 9th

Last Answer : answer:

Solve : (x + 1) (x + 2) (x + 3) (x + 4) + 1 = 0 -Maths 9th

Description : Solve : (x + 1) (x + 2) (x + 3) (x + 4) + 1 = 0 -Maths 9th

Last Answer : answer:

Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Solve for x : log10 [log2 (log39)] = x -Maths 9th

Description : Solve for x : log10 [log2 (log39)] = x -Maths 9th

Last Answer : answer:

Solve log2x + 3 (6x^2 + 23x + 21) = 4 – log3x + 7 (4x^2 + 12x + 9). -Maths 9th

Description : Solve log2x + 3 (6x^2 + 23x + 21) = 4 – log3x + 7 (4x^2 + 12x + 9). -Maths 9th

Last Answer : answer:

Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Description : Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Last Answer : answer:

The ratio of the actual mesh dimension of Taylor series to that of the next smaller screen is (A) 2 (B) √2 (C) 1.5 (D) √3

Description : The ratio of the actual mesh dimension of Taylor series to that of the next smaller screen is (A) 2 (B) √2 (C) 1.5 (D) √3

Last Answer : (B) √2

Sphericity of a cubical particle, when its equivalent diameter is taken as the height of the cube, is (A) 0.5 (B) 1 (C) √2 (D) √3

Description : Sphericity of a cubical particle, when its equivalent diameter is taken as the height of the cube, is (A) 0.5 (B) 1 (C) √2 (D) √3

Last Answer : (B) 1

According to maximum shear stress failure criterion, yielding in material occurs, when the maximum shear stress is equal to __________ times the yield stress. (A) 0.5 (B) 2 (C) √2 (D) √3/2

Description : According to maximum shear stress failure criterion, yielding in material occurs, when the maximum shear stress is equal to __________ times the yield stress. (A) 0.5 (B) 2 (C) √2 (D) √3/2

Last Answer : (A) 0.5

4×(√3 + √4)2 + 6(√5 +√6)2 -3(√2 + √3)2 =? a) 167 b) 123 c) 157 d) 153 e) 149

Description : 4×(√3 + √4)2 + 6(√5 +√6)2 -3(√2 + √3)2 =? a) 167 b) 123 c) 157 d) 153 e) 149

Last Answer : ? =4(3+4+2√12) + 6(5+6+2√30)-3(2+3+2√6) =4(7+2√12)+6(11+2√30)-3(5+2√6) =28 + 8√12 + 66 + 12√30 – 15 - 6√6 =(28+66-15) + (8√12+12√30-6√6) =79 + (8√4 × 3 + 12√30 - 6√6) =79+16√3 + 12√30 - 6√6 =79 + 16 ×1.7 + 12 × 5.4 – 6 ×2.4 =79+27.2 + 64.8-14.4 = 156.6 = 157 Answer: c)

The sum of a positive number and its reciprocal is twice the difference of the number and its reciprocal. The number is : (A) √2 (B) 1/√2 (C) √3 (D) 1/√3

Description : The sum of a positive number and its reciprocal is twice the difference of the number and its reciprocal. The number is : (A) √2 (B) 1/√2 (C) √3 (D) 1/√3

Last Answer : Answer: C

What is a fraction in which the greatest common factor of the numerator and the denominator is 1 is written in what?

Description : What is a fraction in which the greatest common factor of the numerator and the denominator is 1 is written in what?

Last Answer : simplest form

When the numerator and denominator have 1 as a common factor?

Description : When the numerator and denominator have 1 as a common factor?

Last Answer : The fraction is in its simplest form.

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Description : Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Last Answer : Solution: Let p(x) = x3+x2+x+1 The zero of x+1 is -1. [x+1 = 0 means x = -1] p(−1) = (−1)3+(−1)2+(−1)+1 = −1+1−1+1 = 0 ∴By factor theorem, x+1 is a factor of x3+x2+x+1

3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Description : 3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Last Answer : Solution: 7+3x = 0 ⇒ 3x = −7 ⇒ x = -7/3 ∴Remainder: 3(-7/3)3+7(-7/3) = -(343/9)+(-49/3) = (-343-(49)3)/9 = (-343-147)/9 = -490/9 ≠ 0 ∴7+3x is not a factor of 3x3+7x

Show that x + a is a factor of xn + an for any odd +ve integer n. -Maths 9th

Description : Show that x + a is a factor of xn + an for any odd +ve integer n. -Maths 9th

Last Answer : Let f(x) = xn + an x + a will be the factor of xn + an if f(-a) = 0 Now f(-a) = (-a)n + an = 0 (since n is a odd +ve integer) Thus (x +a) is a factor of xn + an .

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Description : If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Last Answer : Since x2 - 1 = (x - 1) is a factor of p(x) = ax4 + bx3 + cx2 + dx + e ∴ p(x) is divisible by (x+1) and (x-1) separately ⇒ p(1) = 0 and p(-1) = 0 p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0 ... (b+d) = 0 ⇒ b + d = 0 ---- (iii) comparing equations (ii) and (iii) , we get a + c + e = b + d = 0

If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.