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Gaurav Seth 5 years, 3 months ago

a n s w e r

Given that :

 Sn  is sum of first  n  terms of an A.P.

Let  a  be the first term and  d  be the common difference of the given  A.P.

Then,

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  • 1 answers

Divyansh Vashishtha 5 years, 3 months ago

2.57577561282
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Gaurav Seth 5 years, 3 months ago

Given quadratic polynomial,
x² - (k+6)x + 2(2k+1) = 0

By comparing it with ax²+bx+c = 0,we get
a = 1 , b = -(k+6) , c = 2(2k+1)

Sum of zeroes = -b/a
= -[-(k+6)]/1
= k+6

Product of zeroes = c/a
= 2(2k+1)/1
= 2(2k+1)

As per problem,
Sum of zeroes = ½ × product of zeroes
k+6 = ½ × 2(2k+1)
k+6 = 2k+1
2k-k = 6-1
k = 5

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Soumyadip Das 5 years, 3 months ago

Geometric Progression

Hema S B 5 years, 3 months ago

Gross product

Ashish V 5 years, 3 months ago

GENERAL PRACTITONER, a doctor who provides general medicine treatment for the people who live in particular area
  • 3 answers

Soumyadip Das 5 years, 3 months ago

10.39

Tapash Das 5 years, 3 months ago

6√3

Seraj Ali ????? 5 years, 3 months ago

6root 3
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Jashraj Bhanushali 5 years, 3 months ago

I know

Gaurav Seth 5 years, 3 months ago

"√5 is an “irrational number”.

Given:

√5

To prove:

√5 is a rational number

Solution:

Let us consider that √5 is a “rational number”.

We were told that the rational numbers will be in the “form” of form Where “p, q” are integers.

So, 

we know that 'p' is a “rational number”. So 5 \times q should be normal as it is equal to p

But it did not happens with √5 because it is “not an integer”

Therefore, p ≠ √5q

This denies that √5 is an “irrational number”

So, our consideration is false and √5 is an “irrational number”."

Seraj Ali ????? 5 years, 3 months ago

Book mai hai bhai
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Mayank Sharma 5 years, 3 months ago

x^2+3x+2 x^2+2x+1x+2 x(x+2)+1(x+2) (x+1)(x+2) x+1=0 x=-1 x+2=0 x=-2
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Chandresh Tiwari 5 years, 3 months ago

6 will not be end with the 0 . Because it does not have factor of 6 ---2 , 5 so it will not end 0.
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Bhumika Kanel 5 years, 3 months ago

Consistent
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Yogita Ingle 5 years, 3 months ago

your question is wrong OK. ? tan48.tan23.tan42.tan67=1 RHS=tan48.tan42.tan23.tan67. =tan48.tan(90-48).tan23.tan(90-23). =tan48.cot48.tan23.cot23. =tan48.1/tan48.tan23.1/tan23 =1 I hope this really help you ???
  • 4 answers

Making Craft Sadaf 5 years, 3 months ago

1/2

Avinash S 5 years, 3 months ago

Sin30=1/2

Richa Shree 5 years, 3 months ago

1/2

Anurag Sharma 5 years, 3 months ago

Testing app?
  • 1 answers

Yadu Raj 5 years, 3 months ago

64x-45y=289 x=289+45y/64......let this be equation 1 Similarly in the second equation 45x-64y=365 x=365+64y/45.......let this be equation 2 From eq 1&2 289+45y/64=365+64y/45 Now by cross multiplication 45(289+45y)=64(365+64y) Simplifying...... 13005+2025y=23360 +4096y Taking LHS to RHS .....we get 2071y=-10355 y=-10355/2071 y=-5 By substituting y in eq 1..... 64x -45 × (-5)=289 64x +225=289 64x=289-225 64x=64 Thefore x=1.......hence x=1 y= -5
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Madhu Sudhan Pandey 5 years, 3 months ago

Yes some questions are come from that concept not this questions come ok
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Yogita Ingle 5 years, 3 months ago

tan 26°/cot 64°   
= tan (90° - 36°)/cot 64°   
= cot 64°/cot 64°
= 1

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Yogita Ingle 5 years, 3 months ago

Let the length of the shorter side b x cm. Then, the length of the longer side = (x + 5) cm.

Since the triangle is right-angled, the sum of the squares of the sides must be equal to the square of the hypotenuse (Pythagoras Theorem).
x2 + (x + 5)2 = 252
or x2 + x2 + 10x + 25 = 625 

or 2x2 + 10x - 600 = 0
or x2 + 5x - 300 = 0 

or (x + 20) (x - 15) = 0

This gives x = 15 or x = - 20 

We reject x = - 20 and take x = 15.
Thus, length of shorter side = 15 cm. 
Length of longer side = (15 + 5) cm, i.e., 20 cm.

 

  • 1 answers

Suraj Verma 5 years, 3 months ago

The nth term of an AP can't be n2+1. Difference between first and second term = 5 - 2 = 3. Their common difference is not equal. In A.P,common difference should not be equal, so this series can't be A.P.
  • 2 answers

Student ✍️✍️✍️ 5 years, 3 months ago

a*3+b*3+3ab(a+b)=a*3+3a*2b+3ab*2+b*3

Reddy Bleela 5 years, 3 months ago

a3+b3+3ab(a+b)

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