CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Left from home and the meet and talk about the day and night and per the meet and talk about the meet only one technique is a joke with the day of any kind of any education and the day is not ?? and per capita income and I am

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Find the zeroes of the polynomial x²+26-25=0

Posted by Nidhi Muliya 4 years ago

CBSE > Class 10 > Mathematics

3 answers

Aadarsh Shrivastav 4 years ago

+,-1

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Shreya Patel 4 years ago

O and 1 are the zeroes of the polynomial

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Zeeshan Zeeshan 4 years ago

X2+26-25=0

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10+6=

Posted by Abhishek Singh 4 years ago

CBSE > Class 10 > Mathematics

5 answers

Vandana Bharti 4 years ago

16?‍♀️

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Naman Gupta 4 years ago

16???

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Komal Kabdal 4 years ago

1Thank You

Himanshu Pandey 4 years ago

16 ?

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Aditi Singhania 4 years ago

16 ??‍♀️

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1 over cosec thitha -1- 1 over cosec thitha +1 =

Posted by Nisarga H 4 years ago

CBSE > Class 10 > Mathematics

0 answers

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The coordinates of mid point of the points (3,5) and (7,9) are

Posted by Gunnu Gungun 4 years ago

CBSE > Class 10 > Mathematics

5 answers

Mahira Srivastava 4 years ago

5,7

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Rifan Rifan 4 years ago

5,7

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Rashmika Singh ?? 4 years ago

5,7

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Ffgh Uigfvb 4 years ago

5,7

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Sumesh Sharma 4 years ago

(5,7)

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The value of (1+tan

Posted by Yuvraj Kumar 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 4 years ago

Please complete your question

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Sia ? 4 years ago

Please complete your q

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If a and b are the zeroes of polynomial f(x)=x2-3x+2 then 1/a and 1/b is equal to

Posted by Sai Ram 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years ago

p(x) = x² - 3x + 2

=> x² - 2x - x + 2 = 0

=> x(x - 2)-1(x - 2) = 0

=> (x - 1)(x - 2) = 0

=> x - 1 = 0, x = 1 => a

=> x - 2 = 0, x = 2 => b

therefore two zeros a and b are 1 and 2

1/a + 1/b = 1/1 + 1/2

=> (2+1)/2

=> 3/2. Answer

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If the circumference of a circle is equal to the perimeter of a square, then what will be the ratio of their areas??

Posted by Fathima Zohra 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Jay Raj Makinidi 4 years ago

14:11

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Anshit Babu 4 years ago

14:11

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Graphically how does the pair of equations 6x-3y+10=0 2x-y+9=0 Represents two lines

Posted by Fathima Zohra 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Jay Raj Makinidi 4 years ago

Parallel lines

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Give the polynomial of degree 2 with sum and product of its zeroes as -1/2 and -3 respectively A) k(x²-x+6) B) k(2x²+x-6) C) k(x²+x-3) D) k(x²-2x+3)

Posted by Fathima Zohra 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Abhi Shek 4 years ago

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If 2 and alpha are zeros of 2x2-6x+2 then value of alpha is

Posted by Tushar Verma 4 years ago

CBSE > Class 10 > Mathematics

1 answers

04 Lily Amitoj Kaur Brar 3 years, 11 months ago

1/2

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If  , are the zeroes of the polynomial 2 f x ax bx c ( ) = + + , then   + = 2 2 1 1 ?

Posted by Itisha Mittal 4 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

You euclid's division Lemma to show that the cube of any positive integers is of the form 9M, 9M + 1 or 9 M + 8

Posted by Pragathi M 4 years ago

CBSE > Class 10 > Mathematics

3 answers

Saumya Pandey 4 years ago

a=bq+r this is a formula

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Aditi Singhania 4 years ago

But whatever the answer is that - An integer can be of the form 9q,9q+1,9q+2,9q+3,.........,or 9q+8

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Aditi Singhania 4 years ago

Pragathi jii don't do the questions related to Euclid's division lemma because it is deleted from the topic in term 1 board examination it is not coming

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If the central of a

Posted by Vinayak Seth 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Srishti Singh 4 years ago

Plzz ask full question

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Sin thetha + cos thetha

Posted by Vishal Kumar 4 years ago

CBSE > Class 10 > Mathematics

4 answers

Sathvik.H Shettigar 4 years ago

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Aditi Singhania 4 years ago

Oh jii the answer is 1

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Vishwajeet Singh 4 years ago

That's sin*2A + cos*2A

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Nikhil Prajapati 4 years ago

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Can any one please tell me the answer ? The zeros of the quadratic polynomial x square+99x+127

Posted by Aditi Singhania 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Yuvraj Agarwal 4 years ago

By direct formula -b/a = -9 / 1

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Anwesha Paul 4 years ago

-1.3 and -97.7

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All class 10th maths formula

Posted by Ashmit Patel 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Rakshit Singh 4 years ago

Book mei dekh le

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Sumit Yadav 4 years ago

Polynomials Formulas ( x + y ) 2 = x 2 + y 2 + 2 x y ( x − y ) 2 = x 2 + y 2 − 2 x y ( x + y ) ( x − y ) = x 2 − y 2 ( x + y ) ( x + z ) = x 2 + x ( y + z ) + y z ( x + y ) ( x − z ) = x 2 + x ( y − z ) − y z x 2 + y 2 = ( x + y ) 2 − 2 x y ( x + y ) 3 = x 3 + y 3 + 3 x y ( x + y ) ( x − y ) 3 = x 3 − y 3 − 3 x y ( x − y ) ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 z x ( x − y − z ) 2 = x 2 + y 2 + z 2 − 2 x y + 2 y z − 2 z x x 3 + y 3 = ( x + y ) ( x 2 − x y + y 2 ) x 3 − y 3 = ( x − y ) ( x 2 + x y + y 2 ) x 4 − y 4 = ( x 2 ) 2 − ( y 2 ) 2 = ( x 2 + y 2 ) ( x 2 − y 2 ) = ( x 2 + y 2 ) ( x + y ) ( x − y ) ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 z x ( x + y − z ) 2 = x 2 + y 2 + z 2 + 2 x y − 2 y z − 2 z x ( x − y + z ) 2 = x 2 + y 2 + z 2 − 2 x y − 2 y z + 2 z x ( x − y − z ) 2 = x 2 + y 2 + z 2 − 2 x y + 2 y z − 2 z x x 3 + y 3 + z 3 − 3 x y z = [ ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) ] 2. Arithmetic Progression Formulas nth Term of an Arithmetic Progression a n = a + ( n − 1 ) × d Sum of 1st n Terms of an Arithmetic Progression S n = n 2 [ 2 a + ( n − 1 ) d ] 3. Coordinate Geometry Formulas Distance Formula A B = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Section Formula ( m x 2 + n x 1 m + n , m y 2 + n y 1 m + n ) Mid-point Formula ( x 1 + x 2 2 , y 1 + y 2 2 ) Area of Triangle ar ( Δ A B C ) = 1 2 × ⎡ ⎢ ⎣ x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) ⎤ ⎥ ⎦ 4. Trigonometry Formulas Trigonometric Identities sin 2 A + cos 2 A = 1 tan 2 A + 1 = sec 2 A cot 2 A + 1 = c o s e c 2 A Relations between Trigonometric Identities tan A = sin A cos A cot A = cos A sin A c o s e c A = 1 sin A sec A = 1 cos A Trigonometric Ratios of Complementary Angles sin ( 90 ∘ − A ) = cos A cos ( 90 ∘ − A ) = sin A tan ( 90 ∘ − A ) = cot A cot ( 90 ∘ − A ) = tan A sec ( 90 ∘ − A ) = c o s e c A c o s e c ( 90 ∘ − A ) = sec A Values of Trigonometric Ratios of 0° and 90° ∠ A 0 ∘ 30 ∘ 45 ∘ 60 ∘ 90 ∘ sin A 0 1 2 1 √ 2 √ 3 2 1 cos A 1 √ 3 2 1 √ 2 1 2 0 tan A 0 1 √ 3 1 √ 3 Not Defined sec A 1 2 √ 3 √ 2 2 Not Defined cosec A Not Defined 2 √ 2 2 √ 3 1 cot A Not Defined √ 3 1 1 √ 3 0 5. Circles Formulas Area of circle π r 2 Diameter of circle 2 r Circumference of circle 2 π r Sector angle of circle θ = ( 180 × l ) ( π r ) Area of the sector = ( θ 2 ) × r 2 Area of the circular ring = π × ( R 2 − r 2 ) θ = Angle between two radii R = Radius of outer circle r = Radius of inner circle 6. Statistics Formulas Mean a m = a 1 + a 2 + a 3 + a 4 4 = n ∑ 0 a n Median M e d i a n = l + ( n 2 − c f f ) h Mode M o = l + ( f 1 − f 0 2 f 1 − f 0 − f 2 ) h 7. Quadratic Equations Formulas Quadratic Equations a x 2 + b x + c = 0 where a ≠ 0 Quadratic Polynomial P ( x ) = a x 2 + b x + c where a ≠ 0 Zeroes of the Polynomial P ( x ) The Roots of the Quadratic Equations are zeroes One Real Root b 2 − 4 a c = 0 Two Distinct Real Roots b 2 − 4 a c > 0 No Real Roots b 2 − 4 a c < 0 8. Triangles Formulas Six elements of triangle Three sides and three angles Angle sum property of triangle Sum of three angles: ∠ A + ∠ B + ∠ C = 180 ∘ Right angled triangle Adjacent Side Opposite Side Hypotenuse Pythagoras Theorem H 2 = A S 2 + O S 2 H = Hypotenuse A S = Adjacent Side O S = Opposite Side Equilateral Triangles All sides are equal Isosceles Triangle Two sides are equal Congruent Triangles Their corresponding parts are equal SSS Congruence of two triangles Three corresponding sides are equal SAS Congruence of two triangles Two corresponding sides and an angle are equal ASA Congruence of two triangles Two corresponding angles and a side are equal 9. Surface Area and Volume Formulas Cuboid Volume of Cuboid (LSA) l × b × h Lateral Surface Area of Cuboid (LSA) 2 h ( l + b ) Total Surface Area of Cuboid (TSA) 2 ( l b + b h + h l ) Cube Volume of Cube x 3 Lateral Surface Area of Cube (LSA) 4 x 2 Total Surface Area of Cube (TSA) 6 x 2 Sphere Volume of Sphere 4 3 × π r 3 Lateral Surface Area of Sphere (LSA) 4 π r 2 Total Surface Area of Sphere (TSA) 4 π r 2 Right Circular Cylinder Volume of Right Circular Cylinder π r 2 h Lateral Surface Area of Right Circular Cylinder (LSA) 2 × ( π r h ) Total Surface Area of Right Circular Cylinder (TSA) 2 π r × ( r + h ) Right Pyramid Volume of Right Pyramid 1 3 × [ Area of the Base ] × h Lateral Surface Area of Right Pyramid (LSA) 1 2 × p × L Total Surface Area of Right Pyramid (TSA) LSA + [ Area of the Base ] Right Circular Cone Volume of Right Circular Cone 1 3 × ( π r 2 h ) Lateral Surface Area of Right Circular Cone (LSA) π r l Total Surface Area of Right Circular Cone (TSA) π r × ( r + L ) Hemisphere Volume of Hemisphere 2 3 × ( π r 3 ) Lateral Surface Area of Hemisphere (LSA) 2 π r 2 Total Surface Area of Hemisphere (TSA) 3 π r 2 Prism Volume of Prism B × h Lateral Surface Area of Prism (LSA) p × h Total Surface Area of Prism (TSA) π × r × ( r + L ) l = Length, h = Height, b = Breadth r = Radius of Sphere L = Slant Height

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If 2 sin2 B -cos2B=2,then B is

Posted by Dichen Sherpa 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Tanya Singh 4 years ago

90° or 180°

1Thank You

Ashmit Patel 4 years ago

Pta nhi

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The perimeter of a sector of a circle with central angle 90° is 25cm. The area of the minor segment of the circle is

Posted by Rimi Das 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Sumit Yadav 4 years ago

Let r be the radius of the circle and θ be the angle. Now, Perimeter of the sector=(2r+2πrθ360) =2r+2×227×r×90360=(2r+11r7)=25r7 Also, 25r7=25⇒r=(25×725)⇒r=7 cm Area of the minor segment=(πr2θ360−12r)2sin θ cm2 =∣∣(227×7×7×90360)−(12×7×7×sin 90°)∣∣ cm2=(772−492) cm2=282 cm2=14 cm2

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in the backyard of house ,Payal

Posted by Meera Wala 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Preeti Dabral 4 years ago

Please complete your question

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solve the following pair of linear equations. (I)ax+by=a-b,bx-ay=a+b

Posted by Sahil Mehra 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Preeti Dabral 4 years ago

ax+by=a-b
ax=a-b-by
x=a-b-by/a←

bx-ay=a+b
substituting
b(a-b-by/a)-ay=a+b
ab-b²-b²y/a-ay=a+b
ab-b²-b²y-a²y/a=a+b
ab-b²-(b²+a²)y=a²+ab
-(b²+a²)y=a²+ab-ab+b²
(b²+a²)y=-(a²+b²)
y=-(a²+b²)/a²+b²
y=-1←

substituting value of y
x=a-b-b(-1)/a
x=a-b+b/a
x=a/a
x=1

0Thank You

Bhawna Yadav 4 years ago

ax+by=a-b ax=a-b-by x=a-b-by/a← bx-ay=a+b substituting b(a-b-by/a)-ay=a+b ab-b²-b²y/a-ay=a+b ab-b²-b²y-a²y/a=a+b ab-b²-(b²+a²)y=a²+ab -(b²+a²)y=a²+ab-ab+b² (b²+a²)y=-(a²+b²) y=-(a²+b²)/a²+b² y=-1← substituting value of y x=a-b-b(-1)/a x=a-b+b/a x=a/a x=1

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In triangle ABC ,right angled at b ,if tan A 4/4 then the value of cosC is

Posted by D Chandini 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Priya Soni 4 years ago

Given that: In ∆ABC, angle b=90° And tan A =4/4=P/B. Find: Cos C=? By using Pythagoras theorem. AC²=AB²+BC² AC²=(4)²+(4)² AC²=16+16 AC²=32 AC²=√32 AC²=⁴√2. Now we find the value of Cos C= B/H= 4/⁴√2= 1/√2. Done??

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The value of k, for which the pair of linear equations kx + y = k2 and x + ky = 1 have infinitely many solutions is.

Posted by Agrim Tripathi 4 years ago

CBSE > Class 10 > Mathematics

2 answers

Preeti Dabral 4 years ago

{tex}\pm 1{/tex}

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Dhairya Sharma? 4 years ago

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ANSWER

if xcosA =8 and 15cosecA=8secA then value of x is

Posted by Abhinav Singh 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Agrim Tripathi 4 years ago

sec A/cosec A=sin A/cos A=tan A=15/8 In a right triangle, a=15x, b=8x thus hypotenuse is c=17x thus cos A=8/17...17.cos A=8....thus, x=8

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I want clear and correct mind map of real numbers and area related to circles

Posted by Blue Heavens Class 10 4 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

In figure (1);(2) ,DE||BC find Ec in(1) and AD in (2)

Posted by Manish Routhan 4 years ago

CBSE > Class 10 > Mathematics

1 answers

Sumit Yadav 4 years ago

In, ΔABC BC || DE In ΔABC and ΔADE ∠ABC = ∠ADE [corresponding angles] ∠ACB = ∠AED [ corresponding angles] ∠A = ∠A common ⇒ ΔABC ~ ΔADE AD/DB = AE/EC 1.5/3 = 1/EC EC = 3 × 1/1.5 EC = 2 cm (ii) In ΔABC and ΔADE ∠ABC = ∠ADE [corresponding angles] ∠ACB = ∠AED [ corresponding angles] ∠A = ∠A common ΔABC ∼ ΔADE AD/DB = AE/EC AD/7.2 = 1.8/5.4 AD = (7.2 × 1.8)/5.4 AD = 2.4 cm

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