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  • 5 answers

Yogita Ingle 6 years, 9 months ago

First term of A.P. = 9 , common difference = 8
if sum of n terms 636,  Sn = (n/2) [ 2×9 + (n-1)8 ] = 636
Sn = n [ 10+ 8n] = 1272
8n2 +10n -1272 = 0
4n2 +5n -636 = 0
By factorising LHS,
(4n+53)(n-12) = 0
hence n = 12

Sankalp Awasthi 6 years, 9 months ago

12

Guddan_ ?? 6 years, 9 months ago

12

Puja Sahoo? 6 years, 9 months ago

12 terms should be taken

Honey ??? 6 years, 9 months ago

12
  • 2 answers

Ritesh Barnwal 6 years, 9 months ago

Vol=64cm^3=> a^3=4^3=>a=4 l=8 b=4 h=4 TSA=2(lb+bh+hl) => 2(32+16+32) => 2(64+16) => 2×80 => 160cm^2

Honey ??? 6 years, 9 months ago

160 cm sq
  • 1 answers

Vaibhav Yadav 6 years, 9 months ago

Multiply (sec theta - tan theta) with (sec theta + tan theta)/(sec theta + tan theta). This will give you x=reciprocal of(sec theta + tan theta). Thus 1/x= sec theta + tan theta
  • 3 answers

Guddan_ ?? 6 years, 9 months ago

13 and 39

Sankalp Awasthi 6 years, 9 months ago

13 and 39

Puja Sahoo? 6 years, 9 months ago

13 and 39.....
  • 3 answers

Honey ??? 6 years, 9 months ago

Simply, do X1+ X3-X2=X4 and same as for Y coordinate

Sankalp Awasthi 6 years, 9 months ago

Ya -3....

Honey ??? 6 years, 9 months ago

-3
  • 2 answers

Muskan Ojha 6 years, 9 months ago

Its*

Muskan Ojha 6 years, 9 months ago

Multiply last by itsome conjicuate
  • 2 answers

Wasim Zafar 6 years, 9 months ago

X-Y=4, x=4+y(1) 3x-2y=10 3(4+y)-2y=10. from1 12+3y-2y=10 Y=-2 From1, X=4+(-2) =2 X=2, y=-2

Ravindranath Bhosale 6 years, 9 months ago

X=4 y=1
  • 2 answers

Yash Sharma ?? 6 years, 9 months ago

14 and 13

Gagan Kaur 6 years, 9 months ago

Let fitst number be x and second be 27+x According to given condition X(27+×) =182 27×+×2 -182 ×2-13×-14×-182 ×(×-13)-14(×-13) (×-13)(×-14) ×=13,×=14
  • 0 answers
  • 1 answers

Pokemon Pro 6 years, 9 months ago

Cos tita is b-a/b
  • 0 answers
  • 4 answers

Affu 😊 6 years, 9 months ago

Yes , it is written in admit card..If you don't have then buy it

Garima Choudhary 6 years, 9 months ago

Yes it is wrritten on admit card

Devansh Rai 6 years, 9 months ago

Yes it is compulsory plus transparent bottle also

Abhishek Singh 6 years, 9 months ago

No
  • 4 answers

Garima Choudhary 6 years, 9 months ago

Ya both are allowed but they both should be transparent
Re nawab mera naam bhi Kabir hai..... Kabir Baisla (Bansal)

Harshita Srivastava 6 years, 9 months ago

Yes but both should be transparent.

#Aditi~ Angel???? 6 years, 9 months ago

No
  • 1 answers

#Aditi~ Angel???? 6 years, 9 months ago

It's given in cbse.nic websites or in anyoth3r websites
  • 1 answers

Vanshika Mittal 6 years, 10 months ago

Cos 60= PR/OR 1/2 = PR/OR OR= 2PR - - - - - - - - - (1) IN TRIANGLE QRO COS60 = RQ/OR 1/2= RQ/OR 2RQ= OR-----------(2) ADDING BOTH, OR+OR = 2PR + 2RQ OR= PR + RQ HENCE PROVED
  • 1 answers

Hiru ? 6 years, 9 months ago

(a-b)(a²+ab+b²)
  • 1 answers

Devansh Rai 6 years, 9 months ago

Ncert solution
  • 1 answers

Sia ? 6 years, 4 months ago

Given, Three cubes of a metal whose edges are in the ratio 3 : 4 : 5{tex}{/tex}.So,we can take the edges of these cubes as  3x, 4x and 5x.
Volume of the cones are, (3x){tex}^3{/tex}, (4x){tex}^3{/tex}, (5x){tex}^3{/tex} i.e  27x{tex}^3{/tex},  64x{tex}^3{/tex},  125x{tex}^3{/tex}.
We know that,
the length of the diagonal of a single cube of side {tex}a = a \sqrt { 3 }{/tex}
The length of the diagonal of a cube is given to be {tex}12 \sqrt { 3 }.{/tex}
So, the edge of a single cube is 12 cm.
Volume of the single cube = (12)3 = 1728 cm3
Since the three cubes are melted and converted into a single cube,
Sum of the volumes of the three cubes = Volume of a single cube
{tex}\Rightarrow{/tex} 27x3 + 64x3 + 125x3 {tex}=1728{/tex}
{tex}\Rightarrow{/tex} 216x3 {tex}= 1728{/tex}
{tex}\Rightarrow{/tex} x3 = 8
{tex}\Rightarrow{/tex} {tex}x = 2 \ cm{/tex}
So, the edges of the cubes are 6 cm, 8 cm and 10 cm.

  • 2 answers

Ishit__ __? 6 years, 10 months ago

mn=(cosecA+cotA)(cosecA-cotA) =>cosec²A-cot²A =>1

Nisha Sandhu (Super Cool) 6 years, 10 months ago

Someone give me answer
  • 1 answers

Sia ? 6 years, 4 months ago

We have,
xsin3{tex}\theta{/tex} + ycos3{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} (xsin{tex}\theta{/tex}) sin2{tex}\theta{/tex} + (ycos{tex}\theta{/tex}) cos2{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} (sin2{tex}\theta{/tex}) + (x sin{tex}\theta{/tex}) cos2{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex} [{tex}\because{/tex} x sin{tex}\theta{/tex} = y cos{tex}\theta{/tex}]
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} (sin2{tex}\theta{/tex} + cos2{tex}\theta{/tex}) = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x = cos{tex}\theta{/tex}
Now, xsin {tex}\theta{/tex} = ycos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} cos{tex}\theta{/tex} sin{tex}\theta{/tex} = y cos{tex}\theta{/tex} [{tex}\because{/tex} x = cos{tex}\theta{/tex}]
{tex}\Rightarrow{/tex} y = sin{tex}\theta{/tex}
Hence, x2 + y2 = cos2{tex}\theta{/tex} + sin2{tex}\theta{/tex} = 1

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