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  • 1 answers

Puja Sahoo 6 years, 11 months ago

If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10. But, 7ⁿ=(7×1)ⁿ Therefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with zero.
  • 2 answers

Gaurav Seth 6 years, 11 months ago

Q. If sec theta = x+1/4x ,prove that :
sec theta +tan theta=2x or 1/2x

<hr />

Solution:

Puja Sahoo 6 years, 11 months ago

Please, check yur question, the value of sec thita given, is wrong according to me...........
  • 3 answers

Puja Sahoo 6 years, 11 months ago

Development refers to the progress, of something........

Kabir Baisla(बागपत) 6 years, 11 months ago

the process of creating something more advanced; a more advanced product...is known as development

Jasline S.? 6 years, 11 months ago

What is development
  • 5 answers

Nikita Sharma 6 years, 11 months ago

1

Puja Sahoo 6 years, 11 months ago

Its....1....?

Kabir Baisla(बागपत) 6 years, 11 months ago

1..................

Diksha Rana 6 years, 11 months ago

1

S.P Singh 6 years, 11 months ago

1
  • 2 answers

Puja Sahoo 6 years, 11 months ago

Ya sure........?

@ Aashu 6 years, 11 months ago

Yup
  • 2 answers

Anjali Kumari??? 6 years, 11 months ago

Do u have any toughest question from chapter light?

Anshika Pal???? 6 years, 11 months ago

376
  • 1 answers

Saisha Sharma 6 years, 11 months ago

By the help of formula of triangle you can find it. X1( Y2 -Y3)+ X2(Y3-Y1)+X3 (Y1-Y2)=0 If theyare collinear
  • 4 answers

Saisha Sharma 6 years, 11 months ago

Yes itis a composite No. 7 (5*1*13+1) 7(65+1) 7×66

Sonu 6 years, 11 months ago

Is divided by 2 . 5...10.. 7.

Sonu 6 years, 11 months ago

Yes

Sonu 6 years, 11 months ago

700
  • 2 answers

Avinash Saigal 6 years, 11 months ago

Radius = 07 cm so, diameter = 7×2 = 14cm the distance between the parallel tangents of circle = diameter of circle = 14 cm Thank you??

Allan Krish 6 years, 11 months ago

14cm
  • 1 answers

Sia ? 6 years, 6 months ago

GIVEN A {tex} \Delta{/tex}DEF and a point X inside it. Point X is joined to the vertices D, E and F. P is any point on DX. {tex} P Q \| D E{/tex} and {tex} Q R \| E F.{/tex}
TO PROVE {tex}P R \| D F{/tex}

CONSTRUCTION Join PR.

PROOF In {tex} \Delta{/tex} XED, we have
{tex} P Q \| D E{/tex}
Therefore,by basic proportionality theorm,we have,
{tex} \therefore \quad \frac { X P } { P D } = \frac { X Q } { Q E }{/tex} ... (i)
In {tex} \Delta{/tex} XEF, we have,
{tex}Q R \| E F{/tex}
Therefore,by basic proportionality theorm,we have,
{tex}\therefore \quad \frac { \mathrm { XQ } } { \mathrm { Q } E } = \frac { \mathrm { X } R } { R F }{/tex} ... (ii) 
From (i) and (ii), we have,
{tex}\frac { X P } { P D } = \frac { X R } { R F }{/tex}
Thus, in {tex}\Delta{/tex} XFD, points R and P are dividing sides XF and XD in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have
{tex} P R \| D F{/tex}

  • 1 answers

_T_A_N_U_78 ??? 6 years, 11 months ago

CSA: pie L(R1+R2) TSA:pieL(R1+R2)+pie R1square+R2square
  • 5 answers

_T_A_N_U_78 ??? 6 years, 11 months ago

OSWAAL CBSE sample question papers,mathematics, class-10

_T_A_N_U_78 ??? 6 years, 11 months ago

No I don't whether this question is from Rs or not l tried to slove this question from oswaal sample paper

Anjali Kumari??? 6 years, 11 months ago

It means this question is from RS?

Anjali Kumari??? 6 years, 11 months ago

24:19 Can u tell me this question is from which book ?

Pankaj Kumar Gupt 6 years, 11 months ago

solution is given in Rs
  • 3 answers

Sahil Rawal 6 years, 11 months ago

First write the sum formula and write the ratio in divide form than you get your answer

Abc . 6 years, 11 months ago

In mean questions first we should take any A(assumed value) from xi (mid value) then ,di=xi-A

Shriya ?? 6 years, 11 months ago

Please tell
  • 2 answers

Riya ? 6 years, 11 months ago

a=bq+r

Abc . 6 years, 11 months ago

a=bq+r Where a=divedened,b=divisor,q=quotient & r is remainder.
  • 1 answers

Gaurav Seth 6 years, 11 months ago

If nth term of an A.P = 2n+1

then, first term of A.P = 2(1)+1
= 2+1
= 3

then, second term = 2 (2) + 1

》4+1

》5


third term = 2 (3) +1

》6+1

》7

sum of terms = 3+5+7

sum of terms = 15

  • 3 answers

Gaurav Seth 6 years, 11 months ago

<header data-role="header" data-sticky="true" id="header">

Let be the first term and d be the common difference of the given AP.

Then,

</header>

Let us now compute,

 

That is the sum of (m +n) terms is zero.

Allan Krish 6 years, 11 months ago

Rs ko pad la Sab question ka answer mil jay gha

Gifty Rose Simon 6 years, 11 months ago

The answer is sloved in r.d sharma
  • 1 answers

Gaurav Seth 6 years, 11 months ago

Let the 1st AP be A =63,65,68.....

Let the 2nd AP be B = 3,7,17......
For A , common difference is : 65-63 = 2 
For B ,common difference is : 10-3 = 7 

Now Let the nth term be An and Bn respectively. 
According to question 
An = Bn 
=> 63+(n-1)2 = 3+(n-1)7
=>60 + 2n-2 = 7n-7
=>65 = 5n 
=> n = 13
Thus 13th term of both AP's will be same 

  • 2 answers

K T 6 years, 11 months ago

My question circumcentre

@ Aashu 6 years, 11 months ago

2pie r
  • 2 answers

Nitya Mishra 6 years, 11 months ago

(5)^2+(x)^2-10x =x^2-10x+25 =x^2-5x-5x+25 =x(x-5)-5(x-5) =(x-5)(x-5) =x=5

Aman Kumar 6 years, 11 months ago

X=5
  • 1 answers

Abhishek Jain 6 years, 11 months ago

Let it is rational and =a/b 2√3=a/b-5
  • 2 answers

Vinit Pandit????? 6 years, 11 months ago

ax+by+c=0 a1x+b1y+c1=0 a/a1=b/b1=c/c1 so =2x+3y-7=0 (m-1)x+(m+1)y=(3m-1) 2/(m-1)=3/(m+1)=7/(3m-1) then M=5answer

Rahul Raj 6 years, 11 months ago

For infinitely many solution a1/a2=b1/b2=c1/c2. Now solve your self
  • 3 answers

Puja Sahoo 6 years, 11 months ago

No, its frm next year

@ Aashu 6 years, 11 months ago

Noo its for coming batch..

Elina ❤️ 6 years, 11 months ago

No......
  • 1 answers

Sia ? 6 years, 4 months ago


Steps of Construction:

  1. Draw a circle with centre O and radius = 4 cm.
  2. Draw any radius OA.
  3. Draw another radius OB such that {tex}\angle{/tex}AOB = 180° - 45° = 135°
  4. At point A draw AP {tex}\bot{/tex} OA.
  5. At point B draw BR {tex}\bot{/tex} OB, intersecting AP at C. AC and BC are required tangents.
  • 1 answers

Sia ? 6 years, 4 months ago

Let the required ratio be k:1.
Then, by the section formula, the coordinates of P are
{tex}P \left( \frac { 9 k + 15 } { k + 1 } , \frac { 20 k + 5 } { k + 1 } \right){/tex}
But, this point is given as P(11, y).
{tex}\therefore{/tex}{tex}\frac { 9 k + 15 } { k + 1 } = 11{/tex} {tex}\Rightarrow{/tex} 9k + 15 = 11k + 11 {tex}\Rightarrow{/tex} 2k = 4 {tex}\Rightarrow{/tex} k = 2
So, the required ratio is 2:1
Putting k = 2 in P, we get
{tex}y = \frac { 20 \times 2 + 5 } { ( 2 + 1 ) } = \frac { 45 } { 3 } = 15{/tex}
Hence, y = 15.

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