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  • 1 answers

My Cbse Guide My Cbse Guide 6 years, 5 months ago

whereis the answer?
  • 0 answers
  • 2 answers

Sunny Kumar 6 years, 5 months ago

10260

Dharmendra Ahirwar 6 years, 5 months ago

10260
  • 1 answers

Shirin Sobnam 6 years, 5 months ago

If the sum of the two angles lying at the transversal is 180° then the angles are co interior angles And the angles must be within the two given parallel lines
  • 3 answers

Rudra Jaggi 6 years, 5 months ago

It resembles identity 1

Shirin Sobnam 6 years, 5 months ago

x^2 + 10x + 25 x^2 + 10x + 5^2 x^2 + 2*x*5 +5^2 (x + 5)^2

Tanishka Kohli 6 years, 5 months ago

(x+5)^2
  • 2 answers

Ranveer Srivastava 6 years, 5 months ago

(a^2 + ab ) + ( ac + bc ) Taking common factors =a(a + b)+c(a + b) ( a + c ) ( a + b ) I hope this will help you .

Tanishka Kohli 6 years, 5 months ago

(a+b)(a+b)
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3/4
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  • 2 answers

Rishita Gupta 6 years, 5 months ago

2x36=72° 3x36=108°

Rishita Gupta 6 years, 5 months ago

2x+3x=180° 5x=180° x=180/5 x=36°
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  • 1 answers

Yogita Ingle 6 years, 5 months ago

first 10 natural number=1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
Mean = 55/10 = 5.5
now, 5th and 6th term as the total number is even .
Median = 5+6/2 = 11/ 2 = 5.5
 

  • 2 answers

Tanishka Kohli 6 years, 5 months ago

0 ans

Akku Kumari 6 years, 5 months ago

a48×48
  • 3 answers

Tanishka Kohli 6 years, 5 months ago

3.1 ,3.2 ,3.3 ,3.4 ,3.5 ,3.6

Kumkum Kumkum 6 years, 5 months ago

Same the denominator by 1 of both the numbers and then add 1 in the no . You want to find like 6+1 and multiply 7 with both the no. Like 3/1×7

Vidya Vishwa1983 6 years, 5 months ago

7/2,11/3,23/6,12/3,25/6,13/6,27/6
  • 3 answers

Santosh Sahu 6 years, 5 months ago

Ok i m

Manoj Kumar 6 years, 5 months ago

No one

Ayush Baranwal 6 years, 5 months ago

I
  • 1 answers

Sia ? 6 years, 5 months ago

It is given that, 
x = 3 + {tex}\sqrt{8}{/tex}
Now, {tex}\frac{1}{x}{/tex} = {tex}\frac{1}{3+\sqrt{8}}{/tex} = {tex}\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}{/tex} ( rationalising the expression) 
{tex}\frac{3-\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}{/tex}
{tex}\frac{3-\sqrt{8}}{9-8}{/tex} = 3 - {tex}\sqrt{8}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{x}{/tex} = 3 - {tex}\sqrt{8}{/tex}
We know that, (x + {tex}\frac{1}{x}{/tex})2 = x2{tex}\frac{1}{x^2}{/tex} + 2 {tex}\times{/tex} x {tex}\times{/tex} {tex}\frac{1}{x}{/tex}
{tex}\Rightarrow{/tex} (3 + {tex}\sqrt{8}{/tex} + 3 - {tex}\sqrt{8}{/tex})2 = x2{tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} (6)2 = x2{tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} 36 = x2{tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} x2{tex}\frac{1}{x^2}{/tex} = 36 - 2 = 34
Hence, x2{tex}\frac{1}{x^2}{/tex} = 34

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Yash Borkar 6 years, 5 months ago

(x+y)square=xsquare +2xy+ysquare

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