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  • 1 answers

Aditi Aditi 4 years, 6 months ago

Its very easy ...You can also find the solution in Rs aggarwal...If the gallery option would have been there I would have sent you
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Atharva Paliwal 4 years, 6 months ago

x^2-2)x^3-3x^2+5x-3(x-3 (-)x^3-2x ―――――――― -3x^2+7x-3 (-)-3x^2+6 ―――――――― 7x-9 Since, degree(r) is less than degree(q) It will not be further divided. For complete division subtract 7x-9 from p(x).......
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Atharva Paliwal 4 years, 6 months ago

Draw the fig.. Join both ends of the design...... Two triangles will form......now, take any quadrant and find its area.Then subtract the area of triangle inside the respective quadrant fro the area of quadrant.. You will find the area of one part of design... Since other quadrant is similar to first one..so double the area of one part of design....you will get the area of whole design........ This was how'll you understand.... For solution:-...2[(area of quadrant)-(area of triangle)]......... May it help....
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A N 4 years, 6 months ago

Regular practice and hard work ; day 1 = ch.1,day 2 = ch.1 +ch. 2 ,day 3= ch.1+ ch.2 + ch.3.........and so on... ;(Ratna nahi hai , concept samajhna hai ) And finally during exam time u need not do all questions. U will automatically b able to answer many questions.

Himanshu Rajput 4 years, 6 months ago

??
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J.Shree Rithanya 4 years, 6 months ago

-b+✓b²-4ac/2a

Priyanka . 4 years, 6 months ago

-b +_ √b²-4ac / 2a

Harsh Nagar 4 years, 6 months ago

-b±√b²-4ac/2a
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Tanvi @1201 4 years, 6 months ago

p(x)=x²-13x+40=0 P(5)=5²-13×5+40=0 =25-65+40=0 =-40+40=0 =0=0 Thus,it is the root of the equation.

Aastha Jha 4 years, 6 months ago

If x= 5 is the root of the equation x² - 13x +40 , then the value of whole equation should be 0....So, Let x=5, then, (5)² - 13*5 + 40 = 0 25-65+40 =0 25-25=0 0=0 Hence, x=5 is the root of the equation.
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Hardik Jain Harsora 4 years, 6 months ago

0.00016

Saliha Tahir 4 years, 6 months ago

First take LCM of 3125 3125=5⁵ Now multiple by 2⁵ (16*2⁵/5⁵*2⁵) 16 *32/10⁵ 512/10⁵ 0.00512 answer
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Mandaloju Smiley 4 years, 6 months ago

18.57
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Punit Patil 4 years, 6 months ago

Wow

Pankaj Singh 4 years, 6 months ago

Nice

Himanshu Rajput 4 years, 6 months ago

Kisss subject me

Himanshu Rajput 4 years, 6 months ago

Weldone??????????????????????????????

Gouri 15 4 years, 6 months ago

Me too in maths
  • 1 answers

Saliha Tahir 4 years, 6 months ago

plz check ur question again I think ur question is wrong...
  • 4 answers

Pankaj Singh 4 years, 6 months ago

All the best

, 🙂 4 years, 6 months ago

All the best

Anil Sah 4 years, 6 months ago

Mere maths ka pre board 2nd March ko tha

Anil Sah 4 years, 6 months ago

Best of luck
  • 5 answers

Ankita Kharkwal 4 years, 6 months ago

Parallel lines

Ravinder Yadav 4 years, 6 months ago

Equation was wrong

Anil Sah 4 years, 6 months ago

Sry parallel line

Anil Sah 4 years, 6 months ago

Intersecting lines

Udit Chauhan 4 years, 6 months ago

"No solution" parallel lines
  • 1 answers

Udit Chauhan 4 years, 6 months ago

TanA•tanB(TanA + tanB)/tanA+tanB
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Kinjal Agarwal 4 years, 6 months ago

Let us considered that 2√3-4 is rational no. That is, it can be written in the form of p/q where p and q are co-primes and q is not equal to 0. Let, "a" be the required rational no. That implies, 2√3-4=a 2√3=a+4 √3=a+4/2 We know that "a" is rational no. That is, a+4/2 is also rational, but √3 is irrational. Which means that, Irrational is not equal to rational Hence, 2√3-4 is irrational Proved!!
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Ankita Kharkwal 4 years, 6 months ago

From RHS ≈2tan theta /1+tan² theta ≈2tan30/1+tan²30 ≈2×1/√3/1+(1/√3)² ≈2/√3/4/3 ≈6/4√3 ≈6/√3/12 ≈√3/2 From LHS Sin2 theta Sin2(30) Sin60 √3/2 LHS =RHS Hence proved

Ayushi Varshney 4 years, 6 months ago

sin2*30°=2tan30°/1+tan30°*tan30°
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Abhijot Bedi 4 years, 6 months ago

No the question is right

Saliha Tahir 4 years, 6 months ago

Take lcm 1/x(a+b+x)=1/ab X a+b-x/x(a+b+x)=a+b/ab X cut by x -a+b/x(a+b+x)=a+b/ab Now-a+b cut by a+b -1/x(a+b+x)=1/ab Cross multiple -ab=x(a+b+x) -ab=ax+bx + x^2 -ab=x(a+b)+x^2 Take ab to right side X^2+(a+b)x+ab=0 By quardic formula -b+/-√b²-4ac/2a (You can see formula from book) Now but values -a-b +/-√(a+b)²+4*1*ab/2 Put identity of (a+b)² -a-b+/- √a²+b²+2ab-4ac/2 -a-b+/-√a² +b²-2ab/2 -a-b +/-√(a-b)²/2 Now take +ve -a-b+a-b/2 A cut by a -2b/2 X= -b Take -ve -a-b-a+b/2 B cut by b -2a/2 X= -a So the values for X = -a , -b

Saliha Tahir 4 years, 6 months ago

If ur question is like this then it is ur ans.... 1/a+b+x=1/a+1/b+1/x Now take 1/x to left side Then , 1/a+b+x-1/x=1/a+1/b Take lcmon both side

Saliha Tahir 4 years, 6 months ago

Because if you don't have x in ur question then how u can solve it for x I think the ques can be like this .... 1/a+b+x=1/a+1/b+1/x

Saliha Tahir 4 years, 6 months ago

Hii I think ur question is wrong
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Shreya Raj 4 years, 6 months ago

It is a branch of mathematics that deals with graph, assumption, probability, and growth rate etc.

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