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Ask QuestionPosted by Ɓӈƛʀƛƭ Rawat 5 years, 2 months ago
- 0 answers
Posted by Ɓӈƛʀƛƭ Rawat 5 years, 2 months ago
- 1 answers
Ɓӈƛʀƛƭ Rawat 5 years, 2 months ago
Posted by Shriram J 5 years, 2 months ago
- 2 answers
Posted by Varad Vitankar 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
Firstly Prepare the proper table of at least two solutions for both the equations..
Draw the graph of both the equations from table 1 & from table 2 on the same graph paper with same scale of representation.
Then find the coordinates of the point of intersection of the graphs.
Here , the two lines intersect at points P (-2,3)
Hence, x= -2 & y= 3
Thus, we have the following table giving points on the line 3y-2y+12 =0
Hence, x =-2 ,y=3 is the solution of the given system of equations.
Posted by Anupama ?? 5 years, 2 months ago
- 2 answers
Posted by Anubhav Thkr 5 years, 2 months ago
- 1 answers
Amrita Kumari 5 years, 2 months ago
Posted by Aniket Bankar 5 years, 2 months ago
- 2 answers
Posted by Nimit Chauhan 5 years, 2 months ago
- 3 answers
Nimit Chauhan 5 years, 2 months ago
Posted by Raman Baghel 5 years, 2 months ago
- 0 answers
Posted by Deepika Tiwari 5 years, 2 months ago
- 2 answers
Yogita Ingle 5 years, 2 months ago
(5+√23) (8-√23)
= 5 (8-√23) + √23 (8-√23)
= 5 (8) - 5 √23 + 8 √23 - √23(√23)
= 40 +(-5 + 8)√23 - 23
= 17 + 3√23
Posted by Shalini Bhadoriya 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
An acute angle of theta is 60 degree.
To find:
Acute angle theta = ?
Solution:
Given:
Applying componendo dividendo, we get
An acute angle is an angle smaller than a right angle. The range of acute angle is one that is less than “90 degrees”. Trigonometric functions of an acute angle are “ratios of the different pairs of sides” of a “right-angled triangle”.
Posted by Deepak Mandal 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
Since, length of two tangents drawn from an external point of circle are equal.
So, AP = AS
BP = BQ DR = DS
and RC = CQ
Adding all, we get
(AP + BP) + (DR + RC) = AS + BQ + DS + CQ
⇒ (AP + BP) + (DR + RC)
= (AS + DS) + (BQ + CQ)
⇒ AB + DC = AD + BC
⇒ 6 + 4 = AD + 7
⇒ 10 = AD + 7
AD = 10 – 7 = 3 cm
Hence, AD = 3 cm.
Posted by Kana Kana 5 years, 2 months ago
- 2 answers
Posted by ?Reetu? A 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
Condition for infinitely many solutions
Since, c has different values.
Hence, for no value of c the pair of equations will have infinitely many solutions.
Posted by Khush Kaur 5 years, 2 months ago
- 5 answers
Rishabh Sharma 5 years, 2 months ago
Posted by Khushi Kachhwaha 5 years, 2 months ago
- 2 answers
Upansh Yadav 5 years, 2 months ago
Posted by Sachin Kumar 5 years, 2 months ago
- 2 answers
Sahiba Tak 5 years, 2 months ago
Posted by Gomsi Sharma 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
In direct variation we do as follows-:
Let the height of the pole be x
6m=4m
x=50m
6/4=x/50. (cross multiply)
50*6=x*4
50*6/4=x
25*3=x
x=75
Means the flag pole height is of 75m.
Posted by Desi Yaari Harayanvi 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
=6x²-7x-3
=6x²+2x-9x-3
=2x(3x+1)-3(3x+1)
=(2x-3)(3x+1)
⇒2x-3=0 or ⇒3x+1=0
⇒x=3/2 or ⇒ x= - 1/3
α=3/2 ,β= - 1/3
⇒α+β= -b/a
⇒3/2+(-1/3)= - (-3)/6
⇒3/2-1/3=1/2
⇒7/6 =1/2
⇒αβ=c/a
⇒ 3/2(-1/3)= -7/6
⇒-1/2= -7/6
⇒1/2=7/6
Posted by Saurabh ???? 5 years, 2 months ago
- 1 answers
Posted by Krish Dedha 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
Sum of roots = -( coeff.of x)/coeff.of x^2
2+3 = -(-2k)/3
5 = 2k/3
2k = 15
k = 15/2
Product of roots = constant term/ coeff.of x^2
2×3 = 2m/3
6 = 2m/3
m = 6 × 3/2 = 9
Posted by Rohit Pal 5 years, 2 months ago
- 1 answers
Harish Sharma 5 years, 2 months ago
Posted by Tanish Kabra 5 years, 2 months ago
- 2 answers
Kiran Kaintura 5 years, 2 months ago
Ashutosh Gupta 5 years, 2 months ago
Posted by Vishakha Arora 5 years, 2 months ago
- 0 answers
Posted by Vedant Patil 5 years, 2 months ago
- 2 answers
Omkar Bhongale 5 years, 2 months ago
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