No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 1 answers

Sia ? 3 years, 2 months ago

(a−2b−3c)2
According to the equation ,
(a+b+c)= a+ b+ c+ 2ab + 2bc + 2ca
So we get,
Using the formula we can it write it as
(a−2b−3c)= a2+(−2b)2+(−3c)2+2a(−2b)+(2(−2b)(−3c)+2(−3c)a
On further calculation we get,
(a−2b−3c)= a+ 4b+ 9c− 4ab + 12bc − 6ac

  • 0 answers
  • 1 answers

Yashika Khantwal 3 years, 2 months ago

Equation is incomplete pls check
  • 0 answers
  • 0 answers
  • 1 answers

Sia ? 3 years, 2 months ago

Given BA || ED and BC || EF.
To show {tex}\angle{/tex}ABC = {tex}\angle{/tex}DEF.
Construction Draw a ray EP opposite to ray ED. 

Proof In the figure, BA || ED or BA || DP
{tex}\therefore{/tex}  {tex}\angle{/tex}ABP = {tex}\angle{/tex}EPC [corresponding angles]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}ABC = {tex}\angle{/tex}EPC ....(i)
Again, BC || EF or PC || EF
{tex}\therefore{/tex} {tex}\angle{/tex}DEF = {tex}\angle{/tex}EPC [corresponding angles] ....(ii)
From eqs.(i) and (ii),
{tex}\angle{/tex}ABC = {tex}\angle{/tex}DEF

X-1
  • 2 answers

Pratham Patel 3 years, 2 months ago

X-1 = X-1

Arman Sheikh 3 years, 2 months ago

VabHa
  • 4 answers

Sia ? 3 years, 2 months ago

(3X + Y)2

Tanu Shri Saini 3 years, 2 months ago

Using identity - x²+2xy+y²=(x+y)² =So, 9x²+6xy+y² =(3x)²+2(3x)(y)+(y)² =(3x+y)²

Tanu Shri Saini 3 years, 2 months ago

Using identity - x²+2xy+y²=(x+y)² 9x²+6xy+y² (3x)²+2(3x)(y)+(y)² (3x+y)² (3x+y) (3x+y)

Pratham Patel 3 years, 2 months ago

(3X+Y)^2
  • 1 answers

Bhoomi Dubey 3 years, 2 months ago

In one variable,there is only y axis(no X axis) . Here y=3 in one variable
  • 5 answers

Harshvivek Lakhera 3 years, 2 months ago

Yes we can write 0 in a form of p/q.

Mithun P 3 years, 2 months ago

Zero can be represented as p/q in the form of 0/1, 0/2, 0/3, 0/4… And so on. Since the number zero divided by any number is equal to zero.

Mithun P 3 years, 2 months ago

Yes ,we can write 0/1 ,0/2 ....

Yashika Khantwal 3 years, 2 months ago

Rational number have 2 terms 1. P/q should be integer 2. Q not be 0 at any case ... Clearly p/q and every rational number not be 0 at any case

Yashika Khantwal 3 years, 2 months ago

No
  • 2 answers

Harshvivek Lakhera 3 years, 2 months ago

Ans-law if exponent shows that M1/2/m1/4=M1/2-1/4. So, 111/2-1/4 1/4 11 is a answer

Yashika Khantwal 3 years, 2 months ago

Ans: Law =a^m /a^n = a^m-n 11^(1/2-1/4) = 11 ^(2-1/4) = 11^(1/4)
  • 1 answers

Bhoomi Dubey 3 years, 2 months ago

72
  • 2 answers

Siddappa Baraddi 3 years, 2 months ago

Pytogorus

Sunny Balhara 3 years, 2 months ago

Zs
  • 1 answers

Saloni P 3 years, 2 months ago

निम्नलिखित संख्याओं को अभाज्य गुणनखंडो के गुणनफल के रूप में धर्म-कर्म कीजिए ‌1.140 2.156 3.3825 4.5005 5.7429
  • 1 answers

Tiksha Sharma 3 years, 2 months ago

Answer
  • 0 answers
  • 1 answers

Rhythm Sadyan 3 years, 2 months ago

X=2/y=3
  • 0 answers
  • 2 answers

Joshi Tha 3 years, 2 months ago

Answer 7x 3 +8y 3 −(4x+3y)(16x 2 −12xy+9y 2 ) =7x 3 +8y 3 −(4x+3y)[(4x) 2 −4x×3y+(3y) 2 ] =7x 3 +8y 3 −(64x 3 +27y 3 ) =−57x 3 −19y 3

Joshi Tha 3 years, 2 months ago

K
  • 1 answers

Joshi Tha 3 years, 2 months ago

Step-by-step explanation: \begin{gathered}Given \: \frac{\sqrt{9}}{27}\\=\frac{\sqrt{3^{2}}}{27}\\=\frac{3}{27}\\=\frac{3}{3\times 9}\\=\frac{1}{9}\\=Rational \:number\end{gathered} Given 27 9 ​ ​ = 27 3 2 ​ ​ = 27 3 ​ = 3×9 3 ​ = 9 1 ​ =Rationalnumber ​
  • 1 answers

Yashveer Choudhary 3 years, 2 months ago

6
  • 0 answers
  • 1 answers

Bhakti Darekar 3 years, 2 months ago

Rroot 6
  • 1 answers

Joshi Tha 3 years, 2 months ago

√6 × √27 The given problem can be solved by expressing each term into individual roots and multiplying √6 can be expressed as √2 × √3 √27 can be expressed as √3 × √3 × √3 = √2 × √3 × √3 × √3 × √3 = 3 × 3√2 = 9√2 Answer √6 × √27=9√2

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App