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  • 4 answers

Shiva Prakash 6 years, 1 month ago

I have read bro

Abhinav ? 6 years, 1 month ago

Me..also..

~ Honey... ? 6 years, 1 month ago

Was u studied?

~ Honey... ? 6 years, 1 month ago

Me
  • 16 answers

~ Honey... ? 6 years, 1 month ago

OK

Lucifer ? Morningstar? 6 years, 1 month ago

Ye chap ke end me h [ frustum ]

~ Honey... ? 6 years, 1 month ago

Ye Formula to nhi padhaya Mujhe abhi

Lucifer ? Morningstar? 6 years, 1 month ago

Ye Monica Capoor ke book me mila

Lucifer ? Morningstar? 6 years, 1 month ago

Volumr of the frustum of the cone=1/3πh(r1^2+r1r2+r2^2). Total surface area=πl(r1+r2)+πr1^2+πr2^2

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of hemisphere=2/3πr^3. Total surface area=3πr^2 Lateral/curved surface area=2πr^2

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of sphere=4/3πr^3. Total surface area=4πr^2. Lateral/curved surface area=4πr^2

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of cone=1/3πr^2h. Total surface area=πr(l+r). Lateral/curved surface area=πrl

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of cylinder=πr^2h. Total surface area=2πr(r+h). Lateral/curved surface area=2πrh

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of cuboid=lbh. Total surface area=2(lb+bh+hl). Lateral surface area=2(l+b)h.

Lucifer ? Morningstar? 6 years, 1 month ago

Volume of cone=a^3. Total surface area=6a^2

Lucifer ? Morningstar? 6 years, 1 month ago

Ok

~ Honey... ? 6 years, 1 month ago

Plz

~ Honey... ? 6 years, 1 month ago

Tell me the formulas pl

~ Honey... ? 6 years, 1 month ago

Will u plz plZ help me

Lucifer ? Morningstar? 6 years, 1 month ago

Yes
  • 2 answers

Rani Mishra ??? 6 years, 1 month ago

You will get in the app

Lalta Ojha 6 years, 1 month ago

I will give you
  • 3 answers

Nandini Kumari Gupta 6 years, 1 month ago

Devansh pandey gave right answer

Devansh Pandey 6 years, 1 month ago

(x + y)³ = x³ + y³ + 3xy ( x + y )

Ankit Kumar 6 years, 1 month ago

(x+y)³
  • 2 answers

Madhu Sharma 6 years, 1 month ago

I think question incomplete h

Devansh Pandey 6 years, 1 month ago

(1+COSA)x(1-cos A)=1-COS²A = sin²A
  • 1 answers

Sahil ??? 6 years, 1 month ago

(5x-8)/(2x^2-9x+9)
  • 1 answers

Jay Kumar Sharma 6 years, 1 month ago

First multiple of 8 is 8 8,16,24...... a is 8 d is 8 Sn =n/2[2a+(n-1)d] 15/2[2*8+(15-1)8] 15/2[16+112] =15*64 =960
  • 0 answers
  • 1 answers

Shubham Kumar 6 years, 1 month ago

We divide X3-3*2+5x+3 by x2-2
  • 0 answers
  • 2 answers

Panchali Pal 6 years, 1 month ago

Median can be found by a method which has a formula... median= l+(n/2-F÷f)h. l is the lower limit of the median class (when class intervals are given) n= summasion of all frquencies F=cumulative frequency of the class preceeding the median class f=frequency of median class and h= class width.

Abhinav ? 6 years, 1 month ago

Median is a measure of central tendency which gives the value of the middle most observation in the data. For finding the median you firstly have to arrange the data in the ascending order and in a frequency distribution table. Then apply the formula: if n is odd then median will be(n+1)/2 th observation. If n is even then median will be average of n/2 and (n/2+1). And the median fpr the grouped data can be find out by : median= l+(n/2-cf/f)×h. Here l =lower limit, n=number of observations, cf= cumulative frequency of class preceding the median class, f= frequency, h = class size.
  • 3 answers

Nandini Kumari Gupta 6 years, 1 month ago

Numerator

Sakshi Jain 6 years, 1 month ago

First take denominator

Naruto Uzumaki??? 6 years, 1 month ago

I think its numerator. But not confirmed still...
  • 1 answers

Sia ? 6 years, 1 month ago

We know that, the lengths of tangents drawn from an external point to a circle are equal.
{tex}\therefore {/tex} TP = TQ
In {tex}\triangle{/tex}TPQ,
TP = TQ
{tex}\Rightarrow{/tex} {tex}\angle{/tex}TQP = {tex}\angle{/tex}TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
{tex}\angle{/tex}TQP + {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Angle sum property)
{tex}\therefore {/tex}2 {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Using(1))
{tex}\Rightarrow{/tex} {tex}\angle{/tex}PTQ = 180º – 2 {tex}\angle{/tex}TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP {tex}\perp{/tex} PT,
{tex}\therefore {/tex} {tex}\angle{/tex}OPT = 90º
{tex}\Rightarrow{/tex}{tex}\angle{/tex}OPQ + {tex}\angle{/tex}TPQ = 90º
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OPQ = 90º – {tex}\angle{/tex}TPQ
{tex}\Rightarrow{/tex} 2{tex}\angle{/tex}OPQ = 2(90º –{tex}\angle{/tex}TPQ) = 180º – 2 {tex}\angle{/tex}TPQ ...(2)
From (1) and (2), we get
{tex}\angle{/tex}PTQ = 2{tex}\angle{/tex}OPQ

  • 1 answers

Sia ? 6 years, 1 month ago

Since, P(E) + P (not E) = 1

P (not E) = 1 - P(E) = 1 - 0.05 = 0.95

  • 0 answers
  • 3 answers

Tannu ? 6 years, 1 month ago

Maths ki book m h dwkh lena

Amirtha Varsha 6 years, 1 month ago

btp

Sidharth ($Id) 6 years, 1 month ago

It states that if a line drawn parallel to a side of traingle which intersects the other 2 sides into 2 distinct points, then the line divides those sides in proportion.
  • 3 answers

Diksha ... 6 years, 1 month ago

Triangles

Sidharth ($Id) 6 years, 1 month ago

Chapter - 6

Ansh Attri 6 years, 1 month ago

Triangles
  • 1 answers

Tannu ? 6 years, 1 month ago

Hmmm

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