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  • 1 answers

Akshat Makhaik 7 years, 3 months ago

a1/a2=b1/b2=c1/c2 (K+1)/5=3k/k=15/5 Cross ✖ a1/a2&b1/b2 We get K(k+1)=5(3k) K²+k=15k K²-14k=0 K-14=0 K=14 Put value of k K+1/5 15/5 3. 3k/k 42/14 =3 Therefore,a1/a2=b1/b2=c1/c2
  • 3 answers

Diwakar Kumar 7 years, 3 months ago

(h)2 = (p)2+(b)2 note:- pythagoras was the student of Thales.

Souravpreet Singh 7 years, 3 months ago

(Hypt.)2=( base)2 +(perp.)2

Swati Bhattacharya 7 years, 3 months ago

Square of Hypotenuse = square of base + square of perpendicular
  • 1 answers

Rajveer Rathod 7 years, 3 months ago

20xⁿ(n=2)-17+3x=2x Therefore,20xⁿ+3x-2x-17=0 Therefore,20xⁿ+x-17=0 Therefore,
  • 1 answers

Rajveer Rathod 7 years, 3 months ago

Simple...... x/x=1 Therefore, 1+1+1+2=5 &1+4=5
  • 2 answers

Diksha Goyal 7 years, 3 months ago

What nonsense is this ' Ayush yelpale' .?

Shivam Yadav 7 years, 3 months ago

15 and 45
  • 2 answers

Rajveer Rathod 7 years, 3 months ago

Its just a way of expressing

Devanshi Grover 7 years, 3 months ago

There is no difference between the two.... Its the same........
  • 0 answers
  • 1 answers

Happy Singh 7 years, 3 months ago

Download 'cbse guide' app for detail syllabus notes of each chapter of all subject
  • 0 answers
  • 1 answers

Sia ? 6 years, 4 months ago


Let height of cone = 2x cm
Radius of cone = 10 cm
In {tex}\triangle{/tex}OAB and {tex}\triangle{/tex}OCD 
{tex}\angle{/tex}AOB = {tex}\angle{/tex}COD [ common ]
{tex}\angle{/tex}OAB = {tex}\angle{/tex}OCD [ each 90o ]
Then, {tex}\triangle{/tex}OAB {tex}\sim{/tex} {tex}\triangle{/tex}OCD [ by AA simiarity ] 
{tex}\therefore{/tex} {tex}\frac { O A } { O C } = \frac { A B } { C D }{/tex} [ corresponding sides of similar triangle are proportional]
{tex}\Rightarrow \quad \frac { X } { 2 X } = \frac { r } { 10 }{/tex}
{tex}\Rightarrow{/tex} r = 5 cm
{tex}\therefore{/tex} {tex}\frac { \text { Volume of small cone } } { \text { Volume of frustum } } = \frac { \frac { 1 } { 3 } \pi ( 5 ) ^ { 2 } \times x } { \frac { 1 } { 3 } \pi ( x ) \left[ 10 ^ { 2 } + 10 \times 5 + 5 ^ { 2 } \right] }{/tex}
{tex}= \frac { 25 x } { x \times 175 }{/tex}
={tex}\frac{1}{7}{/tex}

  • 1 answers

Suvadra Devi Panigrahy 7 years, 3 months ago

(1+√11)/5 or (1-√11)/5
  • 2 answers

Siri.M.V Siri 7 years, 3 months ago

It Is Actually Irrational but ItsValue Is Rational i'e 22/7

Zahira Malik 7 years, 3 months ago

Pie is irrational
  • 6 answers

Disha Rathore 7 years, 3 months ago

2 angles are equal which are opposite to the two given equal sides

Likhitha Jayaram Likhitha 7 years, 3 months ago

2

Harsh Tripathi 7 years, 3 months ago

When two side of triangle are equal then their oppsite angle are equal

Siri.M.V Siri 7 years, 3 months ago

2

Satish Kumar 7 years, 3 months ago

2

Monish Kumar 7 years, 3 months ago

2
  • 1 answers

Sia ? 6 years, 4 months ago

Let us suppose that 'a' be the first term and 'd' be the common difference of the A.P.
Now according to question it is given that

S9 = 162.

Using formula for sum to n terms of A.P, we have
{tex}\Rightarrow \frac{9}{2}\left[ {2a + (9 - 1)d} \right] = 162{/tex}
{tex} \Rightarrow \frac{9}{2}\left[ {2a + 8d} \right] = 162{/tex}
{tex}\Rightarrow{/tex} 9a + 36d = 162 ........................(i)
Let a6 and a13 be the 6th and 13th term of the A.P. respectively.
Therefore, using general form of nth term, we have,a6 = a + 5d and a13 = a + 12d
Since, {tex}\frac{{{a_6}}}{{{a_{13}}}} = \frac{1}{2}{/tex}
{tex} \Rightarrow \frac{{a + 5d}}{{a + 12d}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} 2a + 10d = a + 12d
{tex}\Rightarrow{/tex} a = 2d
Now,substituting the value of a = 2d in equation (i), we get,
9(2d) + 36d = 162
{tex}\Rightarrow{/tex} 18d + 36d = 162
{tex}\Rightarrow{/tex} 54d = 162
{tex}\Rightarrow{/tex} d = 3
{tex}\Rightarrow{/tex} a = 2d = 2 {tex}\times{/tex} 3 = 6 = First term

Therefore, 15th term=a+14d=6+14(3)=48

  • 7 answers

Souravpreet Singh 7 years, 3 months ago

K + cl2 ----> kcl

Varun Srivastav 7 years, 3 months ago

KCl

$@Ny@ R@Jput 7 years, 3 months ago

Kcl

Satish Kumar 7 years, 3 months ago

KCl

Rajesh Kumar Saw 7 years, 3 months ago

Kcl

Mahi Chandel 7 years, 3 months ago

KCl

Rishi Mishra 7 years, 3 months ago

Kcl
  • 1 answers

Mayank Chhabra 7 years, 3 months ago

Just add and than subtract both it. Would become easier for u to solve this
  • 1 answers

Souravpreet Singh 7 years, 3 months ago

cosA+sinA=1
  • 0 answers
  • 1 answers

Rupal ???? 7 years, 3 months ago

Sum = 0,product =-15
  • 1 answers

Sia ? 6 years, 4 months ago

Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a+ 9)x+ 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a2 + 9 = 6a
⇒ a2 - 6a + 9 = 0
⇒ a2 - 3a - 3a + 9 = 0
⇒ a(a - 3)  - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)= 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.

  • 1 answers

Satish Kumar 7 years, 3 months ago

a+7d=a+d/2 2a + 14d = a + d a = -13d a +10d = 1/3(a +3d) + 1 a +10d -1 = 1/3(a + 3d) 3(a + 10d -1) = a +3d 3a + 30d -3 = a+ 3d 2a + 27d =3 2(-13d) + 27d =3(since a = -13d) -26d + 27d = 3 d = 3 a= -13d a = -39 a + 14d = -39 + 42 hence a15 =3
  • 1 answers

Soham Deshpande 7 years, 3 months ago

2x^2-7x+3=0 Divide both sides by 2 2x^2/2-7x/2+3/2=0 x^2-7x/2=-3/2 Third term=(1/2×7/2)^2 (7/4)^2=49/16 Adding third term to both sides x^2-7x/2+49/16=-3/2+49/16 (x+7/4)^2=48/32+98/32 (x+7/4)^2=146/32 x+7/4=+or-under root 146/32 x=-7/4+under root 146/32 OR x=-7/4-under root 146/32.
  • 2 answers

Ajay Raj 7 years, 3 months ago

Where n=40, a=1, and d=1. Therefore answer comes 820

Ajay Raj 7 years, 3 months ago

We can use the formula sum to nth term =n/2{2a+(n-1)d}

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