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  • 2 answers

Shraddha ✨✰✰ 4 years, 11 months ago

,ANSWER 2x+3y=siny 2dx/dx​ + 3dx/dy​=cosy×dx/dy​ 2=(cosy−3)dx/dy​ ⇒dx/dy​=2/cosy−3

~@Kanchi Jawla 4 years, 11 months ago

dy/dx in 2x+3y=siny is given by- 2+3dy/dx=cosy dy/dx dy/dx=2/(cosy-3)
  • 2 answers

Mohit Singh 4 years, 11 months ago

Hii

Shraddha ✨✰✰ 4 years, 11 months ago

What is your question?
  • 2 answers

Gaurav Seth 4 years, 11 months ago

Consider the given events
A = Numbers appearing on two dice are different
B = The sum of the numbers on two dice is 4

Clearly,
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
                     (2, 1), (2, 3), (2, 4), (2, 5), (2, 6),
                     (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
                     (4, 1), (4, 2), (4, 3), (4, 5), (4, 6),
                     (5, 1), (5, 2), (5, 3), (5, 4), (5, 6),
                     (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = {(1, 3), (3, 1) and (2, 2)}

 Now ,

A∩B={(1,3) and (3,1)}

∴ Required probability =P(B/A) = n(A∩B) / n(A) = 2 / 30 =1 / 15

Yogita Ingle 4 years, 11 months ago

When two dice are thrown, there are 36 possibilities.
However, we are given, every time both dice contain different numbers means doublet is not a possibility.
As, number of doublets are 6, so, sample space, S=36-6=30
Number of possibilities of the event when sum of dice is 4, (1,3),(3,1).
Here, we won′t take (2,2) as both dice contain different numbers.So,
E=2
So, P(E)=2/30=1/15

  • 2 answers

Kajal Gupta 4 years, 11 months ago

Thank you, but my doubt is n cap is also a unit vector perpendicular to both a and b vector. Then why can't we use that?

Shraddha ✨✰✰ 4 years, 11 months ago

Cross product of vectors A and B is perpendicular to each vector A and B. ∴ for two vectors →Aand→B if →C is the vector perpendicular to both. =(A2B3−B2A3)ˆi−(A1B3−B1A3)ˆj+(A1B2−B1A2)ˆk .
  • 4 answers

Ghanshyam Deshmukh 4 years, 11 months ago

Maths me question twist krne jaisa kuch nhi hai Basics clear hone chahiye ki question solve kaise krna hai phir toh board waale no. Bas change krte hai

Siddhi Singh 4 years, 11 months ago

Twist

Shraddha ✨✰✰ 4 years, 11 months ago

Twist

Aman Kumar 4 years, 11 months ago

Twist
  • 1 answers
To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system A X = B \displaystyle AX=B AX=B.
  • 1 answers

Gaurav Seth 5 years ago

<th>
Definition A relation is a relationship between sets of values. Or, it is a subset of the Cartesian product A function is a relation in which there is only one output for each input.
Denotation A relation is denoted by “R” A function is denoted by “F” or “f”.
Example R = {(2, x), (9, y), (2, z)}

 

** It is not a function, as “2” is input for both x and z.

F = {(2, x), (9, y), (5, x)}
Note: Every relation is not a function. Every function is a relation
  • 2 answers

Suryakant Swain 4 years, 11 months ago

2
x² + 3x + 2 Ans.  Let y = x² + 3x + 2 .•.dy/dx = 2x + 3×1 + 0 = 2x + 3 →d²y/dx² = d/dx.(dy/dx) = 2 × 1 + 0 = 2
  • 2 answers
9999999999999999999999999999999999999999999999999999999999999999999999999.................... To +∞

Prem Joshi 4 years, 11 months ago

9999
  • 1 answers

Palak Sharma 5 years ago

-sinx/1+cosx
  • 2 answers

Mohsin Raza 4 years, 11 months ago

Sec^2 root x /2 root x

Palak Sharma 5 years ago

Logcos(root x) .1/2rootx
  • 1 answers

Gaurav Seth 5 years ago

1. Understand the contents and weightage of the syllabus

2. Preparation from NCERT textbooks is the most important part

Though supplementary books may be used as references, the NCERT textbooks offer an exhaustive range of exercise questions and solved examples on every topic, which is more than enough to do well in your board examinations.

3. Strategize and THEN prepare!

The long form questions (5-6 marks), which are the most feared aspect of a paper, usually come from one of the following sections:

  • Calculus, which carries a whopping 44% weightage, and can be really scoring
  • Differential Equations consisting of the application based difficult questions.
  • Vectors and 3D geometry carry the next highest 17% weightage.

The aforementioned topics are practice-based and the best way to gain perfection in them is to practice and solve more and more problems on them. Then,

  • Probability, Relations and Functions and Algebra have a weightage of 10%, 10% and 13% respectively.
  • Linear Programming: 6%

These topics don’t require additional practice beyond the ones you have covered in your NCERT textbooks.

In linear programming, a 5-mark question comes from this chapter, which can be dealt with through little practice of understanding the questions and writing the linear equations.

4. Practice sample papers and previous years’ papers

5. Say goodbye to rote learning!

6. Presentation is very important

7. Time management

  • 5 answers

Krati Varshney 5 years ago

7.7 and 7.8 cut ho gyi h according to cbse deleted syllabus...
Sab ko pahle to thanks & sorry bcz mene aap logo ko paresan kiya puch k , but me isliye puch kyuki hamare school ka syllabus R.O. pune se decide hota hai. To koi hamare pune reason ka hota to muje syllabus ka completely pata chal jata.

Gursimran Singh 5 years ago

Hamare toe pura integral aa raha hai

Mishti ???? 5 years ago

It is made separately by schools ... Whatever syllabus u r school is decided to give it will came in u'r exam dear...
It is based on your school teacher.
  • 1 answers

Sweety Kadyan 5 years ago

Yes, first give priority to NCERT and then prefer extra books .Doing NCERT two times and try to solve last year question and sample paper will help u a lot to score goodmarks in your board exam. All the best dear ..
  • 1 answers

Sweety Kadyan 5 years ago

Yes , but also try to do last year question paper that will make u ensure about Ur preparation level and u will get familiar with question paper patter
  • 1 answers

Divyansh Gupta 4 years, 11 months ago

׳/4
  • 2 answers

Dp Gour 5 years ago

AB=I(identity fn.)
AB = BA =I the relation of an inverse of a matrix is for any two matrices A & B , the inverse is only possible if AB = BA = I so it may apply conversely on your question Hope it may help
  • 2 answers
Sorry ans . adhura chhod diya. Then, dv/dx = 1/√1-(x)².d√x/dx , dv/dx =1/√1-x.1/2√x. , dv/dx = 1/2√x×√1-x. , dv/dx = 1/2√x(1-x). , dv/dx = 1/2√x-x² Then , dy/dx = du/dx + dv/dx , then dy/dx = x^sinx ×(sinx + x cosx .logx )/x + 1/2√x-x² , and first vale ans me jo dy/dx find kiya hai vo du/dx hai. Ok
dy/dx = du/dx. +. dv/dx then, first we find the du/dx y = x^sinx then , y = e^sinx.logx Differ. W.r.t. x dy/dx = d/dx (e^ sinx.logx) dy/dx =e^sinx.logx .d/ dx(sinx.logx) dy/dx = x^sinx [sinx .d/dx logx + logx.d/dx sinx ] dy/dx = x^sinx (sinx + x cosx.logx )/x then , we find dv/dx dv/dx = d(sin-¹√x)/dx dv/dx = 1/√(1
  • 1 answers

Shivam Nosaliya 5 years ago

-3 8 -8 5
  • 2 answers

Akanksha Kumari 5 years ago

Thanks yrr

Unnat Mishra 5 years ago

Let the side of cube be x Volume of cube=x^3 V=x^3 Differentiating both side with respect to t dV/dt=3x^2 dx/dt = k Where k=constant dx/dt= k/3x^2 ... (1) Surface area = 6x^2 Differentiating both side dS/dt = 12x dx/dt dS/dt = 12x k/3x^2 (from 1) dS/dt = 4k/x dS/dt is directly proportional to 1/x Thus, the increase in surface area varies inversally as the length of side
  • 4 answers

Arjit Arjit 5 years ago

Here let infinite be a number and let it be £ Now infinite×infinte gives Square of infinite Which is belongs to infinite Hence our assumption is wrong infinite is not a number Thus infinite×infinite is also infinite

Ayush Mohite 5 years ago

You can say infinite square but that will be once more infinite in this case since the answer infinite is greater than the product of two infinites we can get that infinite 1 (the answer) >> infinite 0 (the products)

Arpit Namdev 5 years ago

Its an undeterminant form

Chandan Jaiswal 5 years ago

You will get infinite.
  • 2 answers

Chandan Jaiswal 5 years ago

For 12 elements possible order are 1x12, 12x1, 2x6, 6x2, 3x4,and 4x3 For 7 elements possible order are 7x1, 1x7

Yogita Ingle 5 years ago

for 12 elements , possible orders are 1 × 12, 2 × 6, 3 × 4 , 4 × 3 and 12 × 1.
for 7 elts. possible orders are only 1 × 7 and 7 × 1

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