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  • 4 answers

Shiva Prakash 6 years, 5 months ago

I have read bro

Abhinav ? 6 years, 5 months ago

Me..also..

~ Honey... ? 6 years, 5 months ago

Was u studied?

~ Honey... ? 6 years, 5 months ago

Me
  • 16 answers

~ Honey... ? 6 years, 5 months ago

OK

Lucifer ? Morningstar? 6 years, 5 months ago

Ye chap ke end me h [ frustum ]

~ Honey... ? 6 years, 5 months ago

Ye Formula to nhi padhaya Mujhe abhi

Lucifer ? Morningstar? 6 years, 5 months ago

Ye Monica Capoor ke book me mila

Lucifer ? Morningstar? 6 years, 5 months 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, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

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

Lucifer ? Morningstar? 6 years, 5 months ago

Ok

~ Honey... ? 6 years, 5 months ago

Plz

~ Honey... ? 6 years, 5 months ago

Tell me the formulas pl

~ Honey... ? 6 years, 5 months ago

Will u plz plZ help me

Lucifer ? Morningstar? 6 years, 5 months ago

Yes
  • 2 answers

Rani Mishra ??? 6 years, 5 months ago

You will get in the app

Lalta Ojha 6 years, 5 months ago

I will give you
  • 3 answers

Nandini Kumari Gupta 6 years, 5 months ago

Devansh pandey gave right answer

Devansh Pandey 6 years, 5 months ago

(x + y)³ = x³ + y³ + 3xy ( x + y )

Ankit Kumar 6 years, 5 months ago

(x+y)³
  • 2 answers

Madhu Sharma 6 years, 5 months ago

I think question incomplete h

Devansh Pandey 6 years, 5 months ago

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

Sahil ??? 6 years, 5 months ago

(5x-8)/(2x^2-9x+9)
  • 1 answers

Jay Kumar Sharma 6 years, 5 months 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, 5 months ago

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

Panchali Pal 6 years, 5 months 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, 5 months 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, 5 months ago

Numerator

Sakshi Jain 6 years, 5 months ago

First take denominator

Naruto Uzumaki??? 6 years, 5 months ago

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

Sia ? 6 years, 5 months 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, 5 months 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, 5 months ago

Maths ki book m h dwkh lena

Amirtha Varsha 6 years, 5 months ago

btp

Sidharth ($Id) 6 years, 5 months 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, 5 months ago

Triangles

Sidharth ($Id) 6 years, 5 months ago

Chapter - 6

Ansh Attri 6 years, 5 months ago

Triangles
  • 1 answers

Tannu ? 6 years, 5 months ago

Hmmm

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