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Verify the 3,-1 and -1/3 or the zeros of the cubic polynomial p(x)3=3x3-5x2-11x and verify the relation between its -3 zeros and co efficients

Posted by 10D 12 Siddharth .S 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Aditi Aditi 5 years 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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Divide: (x3-3x2+5x-3) by (x2-2)

Posted by Serma Kani 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Atharva Paliwal 5 years 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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Chapter 5 ex 5.3 question 20

Posted by Khushboo Farooq 5 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

In figure 8.106, a circle is inscribed in AABC having sides AB 8 cm, BC = 7 cm and AC=5.Find AD, BE and CE

Posted by Tanishk 5 years ago

CBSE > Class 10 > Mathematics

0 answers

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Give me Chapter 13 formules

Posted by Azzu Pgl 5 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Calculate the area of the designed region in common between the two quadrants of circles of radius 8 CM each.

Posted by Akansha Garg 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Atharva Paliwal 5 years 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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Yah maths ?Bhi Kya Bala Hai Kabhi dimag Mein Goshti? hi nahin hai ?Koi tips de do naa?

Posted by Shruti Kushwaha 5 years ago

CBSE > Class 10 > Mathematics

2 answers

A N 5 years 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.

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Himanshu Rajput 5 years ago

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1) There is a water tank in the shape of cube having a side of length 4 cm. find the lateral surface, total surface area And find the how much litre of water contain . 2) a room is in the form of cuboid having length 5cm b=4cm if the rate of painting of a wall is 20 Rs per m/sq find the total cost of painting of the four walls (Please friends help me to solve this question)

Posted by Ambika Ambika 5 years ago

CBSE > Class 10 > Mathematics

2 answers

Ambika Ambika 5 years ago

Thank you Himhanshu

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Himanshu Sharma 5 years ago

As weekend know , that LSA of Cube = 4× side × side = 4×4×4 = 64 TSA of Cube = 6× side × side = 6×4×4 = 96 Volume of Cube = side × side × side = 4 × 4 × 4 = 64

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ps app ho

Posted by Himanshu Rajput 5 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Check wheather the pair of linear equation x +3 y -6=0 and 2x + 17 y - 13 =0 is consistent

Posted by Arpita Pathak 5 years ago

CBSE > Class 10 > Mathematics

2 answers

Mayank Choudhary 5 years ago

Consistent

4Thank You

Arya Deo X B 5 years ago

Consistent

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Root 5 is a rationla number prove that

Posted by Suman Sangam 5 years ago

CBSE > Class 10 > Mathematics

0 answers

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Sin theta - cos theta +1 / sin theta + cos theta - 1 = 1/ sec theta - tan theta

Posted by Priyanka . 5 years ago

CBSE > Class 10 > Mathematics

0 answers

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Quadritac formula

Posted by Ronak Sharma.. 5 years ago

CBSE > Class 10 > Mathematics

3 answers

J.Shree Rithanya 5 years ago

-b+✓b²-4ac/2a

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Priyanka . 5 years ago

-b +_ √b²-4ac / 2a

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Harsh Nagar 5 years ago

-b±√b²-4ac/2a

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If Tan A/ Tan B = n and Sin A / Sin B = m. Then show that cos A =√(mn) ^2+m^2-n^2-1/n^4-1

Posted by Aarush Awasthi 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Ronak Sharma.. 5 years ago

Okk

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Find whether x=5 is the root of the equation x² -13x +40=0

Posted by Punit Patil 5 years ago

CBSE > Class 10 > Mathematics

2 answers

Tanvi @1201 5 years 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.

1Thank You

Aastha Jha 5 years 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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For what value of linear equation does has infinitly many solutions of equation ax-3y =1?

Posted by Ishwari Santosh Lamkhade Lamkhade 5 years ago

CBSE > Class 10 > Mathematics

0 answers

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Write down the decimal expansion of 16/3125

Posted by Shree Raksha 5 years ago

CBSE > Class 10 > Mathematics

2 answers

Hardik Jain Harsora 5 years ago

0.00016

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Saliha Tahir 5 years 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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√345

Posted by Ayushi Dubey 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Mandaloju Smiley 5 years ago

18.57

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I have got 80 / 80 in my preboard ?

Posted by Shreya Raj 5 years ago

CBSE > Class 10 > Mathematics

5 answers

Punit Patil 5 years ago

Wow

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Pankaj Singh 5 years ago

Nice

1Thank You

Himanshu Rajput 5 years ago

Kisss subject me

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Himanshu Rajput 5 years ago

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

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Gouri 15 5 years ago

Me too in maths

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1) in the arcs are drawn by taking vertices A,B, and C of an equilateral triangle of side 10cm to intersect the sides BC,CA and AB at their respective midpoints D,E, and F find the area of the shaded region (use π= 3.14) 2) find the perimeter of the shaded region of ABCD is a surface of side 21cm and APB and CPD are semicircle .

Posted by Ambika Ambika 5 years ago

CBSE > Class 10 > Mathematics

0 answers

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(1+cotA +tanA)(sinA- cosA)=secA/cosec^2A- cosecA/ sec^2A = sinAtanA - cotAcosA

Posted by Saanvi Sharma 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Saliha Tahir 5 years ago

plz check ur question again I think ur question is wrong...

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?? ???? ???ℎ ?? ???-????? ℎ??

Posted by Himanshu Rajput 5 years ago

CBSE > Class 10 > Mathematics

4 answers

Pankaj Singh 5 years ago

All the best

0Thank You

, 🙂 5 years ago

All the best

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Anil Sah 5 years ago

Mere maths ka pre board 2nd March ko tha

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Anil Sah 5 years ago

Best of luck

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which type of lines are represented by pairs of equations 5x - y + 6 = 0 , 2x - 2/5 + 7= 0

Posted by R S 5 years ago

CBSE > Class 10 > Mathematics

5 answers

Ankita Kharkwal 5 years ago

Parallel lines

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Ravinder Yadav 5 years ago

Equation was wrong

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Anil Sah 5 years ago

Sry parallel line

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Anil Sah 5 years ago

Intersecting lines

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Udit Chauhan 5 years ago

"No solution" parallel lines

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Prove that: tanA+tanB/cotA+cotB = tanA.tanB

Posted by Om Kumar 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Udit Chauhan 5 years ago

TanA•tanB(TanA + tanB)/tanA+tanB

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Prove that 2underroot 3 - 4 is an irrational no

Posted by Sweety Singh 5 years ago

CBSE > Class 10 > Mathematics

1 answers

Kinjal Agarwal 5 years 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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For θ=30°, verify that : Sin2θ=2tanθ/1+tan²θ

Posted by Om Kumar 5 years ago

CBSE > Class 10 > Mathematics

2 answers

Ankita Kharkwal 5 years 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

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Ayushi Varshney 5 years ago

sin2*30°=2tan30°/1+tan30°*tan30°

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1/a+b+c=1/a+1/b+1/c .....solve for x

Posted by Abhijot Bedi 5 years, 1 month ago

CBSE > Class 10 > Mathematics

5 answers

Abhijot Bedi 5 years ago

No the question is right

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Saliha Tahir 5 years, 1 month 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

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Saliha Tahir 5 years, 1 month 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

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Saliha Tahir 5 years, 1 month 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

1Thank You

Saliha Tahir 5 years, 1 month ago

Hii I think ur question is wrong

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ANSWER

If A>B then sinA >sinB