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

X²-√3² using identity (a+b) (a-b) =a²-b² a=x , b=√3 Putting value in identity (X+√3) (x-√3) =x²-√3² X+√3=0 X=-√3 X-√3=0 X=√3 Zeroes are -√3, √3 Hope you like this

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In triangle ABC,DE perpendicular to BC and angle d is equal to 52 degree ,AD/DE=AE/EC ,find angle ABC

Posted by Lululandia Kalagandya 5 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Waxy Panda 5 years, 3 months ago

52°

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Aashish Kumar 5 years, 3 months ago

Frec

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Find the roots of the following quadratic equations. By factorisation x²-3x-10=0

Posted by Ritika Bansal 5 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Lifes Crush 5 years, 3 months ago

Gaurav, you always help everyone. I want to know your class. Can you please....

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Gaurav Seth 5 years, 3 months ago

x² – 3x – 10
= x² - 5x + 2x - 10
= x(x - 5) + 2(x - 5)
= (x - 5)(x + 2)
Roots of this equation are the values for which (x - 5)(x + 2) = 0

∴ x - 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

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5 tan² A – 5 sec² A + 1 is equal to

Posted by Gkss C Sec 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Santosh Vijay 5 years, 1 month ago

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9. If the pth, qth&rth term of an AP is x, y and z respectively, show that x(q-r) + y(r-p) + z(p-q) = 0

Posted by Albert Einstein 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Khushi Ramchandani 5 years, 3 months ago

Put the values of p , q and r in the equation given , then you will get the answer i.e. LHS =RHS

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X2+5x+10=0

Posted by Reshank Kaiwart 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Palak Gupta 5 years, 3 months ago

Using quadratic formula X= -b±√b²-4ac X= -5± √(5) ²-4*1*10 X= -5±√25-40 X= -5± √15 X= -5±3.8 By taking (+) X= -5+3.8 X= -1. 2 By taking (-) X= -5-3.8 X= -8. 8

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Sides AB and AC and median AD of a triangle ABC are Proportional to sides PQ & PR & median PM of another triangle PQR .Show that triangle ABC~triangle PQR. Please solve in short

Posted by Prajnasree Behera 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Khushi Ramchandani 5 years, 3 months ago

AB=PQ ( sides of the same triangle) AC=PR (sides of the same triangle) BC=QR (sides of the same triangle) Therefore, ABC congruence to PQR ( by sss test )

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-√2 and √2 quadratic equation

Posted by Vikas Ghoderao 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Anurag Prajapati 5 years, 3 months ago

P(x)=x²-(Sum of zeroes)x+(product of zeroes). Put these values in the bracket as given by question and you will get the correct quadratic equation.

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Qutratic polynomial 1/3 and 1/4

Posted by Vikas Ghoderao 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Anurag Prajapati 5 years, 3 months ago

P(x)=x²-(Sum of zeroes)x+(product of zeroes). Put these values in the bracket as given by question and you will get the correct quadratic equation.

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Find the area of a quadarant of a Circle whose circumference 22cm

Posted by Manjula Desai 5 years, 3 months ago

CBSE > Class 10 > Mathematics

3 answers

Vignesh Babu 5 years, 3 months ago

????????????? = ???? ?π? = ?? 2 x 22/7 x r = 22 r = 22 x 7/2 x 22 ? = ?/? In quadrant, angle of sector, = 90° Area of the quadrant = 90°/360° × πr^2 = 40 × 22/7 × 7/2 × 7/2 = 1540 cm^2 ᎻᎾᏢᎬ ᎷᎽ ᎯᏁᏕᏯᎬᏒ ᏯᎨᏝᏝ ᏰᎬ ᎻᎬᏝᏢƒυℓℓ

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Muskan Kumari 5 years, 3 months ago

Circumference= 2×22/7×r 22=2×22/7×r r=22×7/2×22 r=7/2=3.5 cm Area of quadrant= 1/4×22/7×7/2×7/2 =77/8cm²

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Gaurav Seth 5 years, 3 months ago

Formulas used =
1. Circumference = 2 pie r
2. Area of Quadrant = 1/4 Pie r²

Circumference = 2 Pie r
22 = 2 * 22/7 r
r = 22 * 7 / 22 * 2
r = 7/2 or 3.5 cm

Area Of Quadrant = 1/4 Pie r²
= (1/4) * (22/7 ) * 3.5 * 3.5
= 1 * 22 * 35 * 35 / 4 * 7 * 10 * 10
= 1 * 11 * 35 / 4 * 10
= 9.625 cm²

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X square - 9 /x power 4 + x power 3 - 23x square - 3x + 60

Posted by Aditya Kumar 5 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

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Solve for x and y x+6/y=6 3x-8/y=5

Posted by Divyanshi Rathore ❣ 5 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Siba Narayana Dash 5 years, 3 months ago

Above Answer is wrong ...

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Gaurav Seth 5 years, 3 months ago

The given equations are: x + 6/y = 6 ……..(i)

3x - 8/y = 5 ……..(ii)

Putting 1/y = v, we get:

x + 6v = 6 …….(iii)

3x – 8v = 5 ……(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:

4x + 24v = 24 ……..(v)

9x – 24v = 15 ……..(vi)

On adding (v) from (vi), we get:

13x = 39

⇒ x = 3

On substituting x = 3 in (i), we get:

3 + 6/y = 6

⇒ 6/y = (6 – 3) = 3

⇒ 3y = 6

⇒ y = 2

Hence, the required solution is x = 3 and y = 2.

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Solve for x and y x+1/2 +y-1/3=8 x-1/ 3 +y +1 /2=9

Posted by Divyanshi Rathore ❣ 5 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

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What is quadratic polynomial?

Posted by Nagaraj Km 5 years, 3 months ago

CBSE > Class 10 > Mathematics

4 answers

Piyush Agarwal 5 years, 3 months ago

Polynomial in the form of ax*2+bx+c

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Anupama ?? 5 years, 3 months ago

Polynomial having the power 2

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Anupama ?? 5 years, 3 months ago

The equation in the form of x2-ax-b

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Roshesh Tembhurkar 5 years, 3 months ago

A polynomial in which the power is 2

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Find the zeroes of the given quadratic polynomial and verify the relationship between the zeroes and the coefficients: 3x2 – x 4.

Posted by Mathew L Rera 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years, 3 months ago

Since we have given that

First we will find the zeroes of the quadratic polynomial.

We will use "Split the middle terms":

Now, Let,

Now, we will verify the relationship between the zeroes and coefficient.

Sum of zeroes is given by

Hence, verified.

Roots of quadratic polynomials are

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Example 6 of chapter 9

Posted by Tej Pratap Singh 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Ankita Sah 5 years, 3 months ago

In ∆ PBD Tan30°=PD/BD 1/√3=PD/BD BD=PD√3____________by eq .1 In∆PAC Tan45°=PC/AC 1=PC/AC PC=PD+AC PD+DC=AC

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Find the area of a triangle formed by joining the midpoints of the side of the triangle whose vertices are (0,-1),(2,1) and (0,3). Find the ratio of the area of the triangle formed to the area of the given triangle.

Posted by Unknown Girl 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Tej Pratap Singh 5 years, 3 months ago

Out of syllabus

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Ncert ex-13.2 page no.248 question number 3...

Posted by #?Abhishek...? . 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Prajnasree Behera 5 years, 3 months ago

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends. So, the total height of a gulab jamun = 5 cm. Diameter = 2.8 cm So, radius = 1.4 cm ∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm =2.2 cm Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres = πr2h+(4/3)πr3 = 4.312π+(10.976/3) π = 25.05 cm3 We know that the volume of sugar syrup = 30% of total volume So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3) = 45×7.515 = 338.184 cm3

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Solve the quadratic equation in variable x : abx² = (a+b)²(x-1)

Posted by Gargi Nawghade 5 years, 3 months ago

CBSE > Class 10 > Mathematics

4 answers

Gargi Nawghade 5 years, 3 months ago

I hope you understand

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Gargi Nawghade 5 years, 3 months ago

Ans: abx²=(a+b)²(x-1) abx²=(a+b)²x-(a+b)² abx²-(a+b)²x+(a+b)²=0 here, a=ab , b=-(a+b)² , c=(a+b)² By quadratic formula x=-b +/- _/b²-4ac /2a = -[-(a+b)²]+/- _/ [-(a+b)²]²-4(ab)(a+b)²] / 2ab =(a+b)²+/- _/(a+b)⁴-4ab(a+b)² / 2ab = (a+b)²+/- _/(a+b)²[(a+b)²-4ab] / 2ab = (a+b)²+/- (a+b)_/a²+b²+2ab-4ab /2ab = (a+b)²+/-(a+b)_/a²+b²-2ab / 2ab = (a+b)²+/-(a+b)_/(a-b)² / 2ab =(a+b)²+/-(a+b)(a-b) / 2ab = a²+b²+2ab +/- a2-b² /2ab So, x=a²+b²+2ab+a²-b² / 2ab = 2a²+2ab /2ab = 2a(a+b) /2ab = a+b/b OR x=a²+b²+2ab-(a²-b²) / 2ab =a²+b²+2ab-a²+b² / 2ab = 2b²+2ab / 2ab =2b(b+a) /2ab = a+b/a Thus, the value of x= a+b/b , a+b/a

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Tejaswini Chandok 5 years, 3 months ago

Boubht baar try kara meine

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Tejaswini Chandok 5 years, 3 months ago

Nhi hua yr

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it is the given that triangle ABC congruent triangle d e f equal to 1 upon 5 then the formula is does not pass is equal to (a) 25 (b)5 (c)1/25 (d) 1/5

Posted by Vansh Gupta 5 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

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tanq +cotq=4 then the value of the tan^4q+ cot^4q=

Posted by Amit Chaudhary 5 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Tejaswini Chandok 5 years, 3 months ago

Given Equation is tanA + cotA = 4. On squaring both sides, we get = > (tanA + cotA)^2 = (4)^2 = > tan^2A + cot^2A + 2tanAcotA = 16 = > tan^2A + cot^2A + 2 * tanA * (1/tanA) = 16 = > tan^2A + cot^2A + 2 = 16 = > tan^2A + cot^2A = 16 - 2 = > tan^2A + cot^2A = 14. On squaring both sides, we get = > (tan^2A + cot^2A)^2 = (14)^2 = > tan^4A + cot^4A + 2 * tan^4A * cot^4a = 196 = > tan^4A + cot^4A + 2 * tan^4A * (1/tan^4A) = 196 = > tan^4A + cot^4A + 2 = 196 = > tan^4A + cot^4A = 196 - 2 = > tan^4A + cot^4A = 194.

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Vansh Gupta 5 years, 3 months ago

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In a cyclic quadrilateral bacd , angle a =2x-4 degree, angle b= y+5 degree , angle c =2y+10 degree and angle d= 4x-2 degree find the four angles

Posted by Mega K 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Vikarn Jha 5 years, 3 months ago

Chapter linear equation in two varaiable

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The cuboid tent has a square base of side x meters and height h meters. The company uses 8m2 of material to make each cuboid tent. What is the volume of the cuboid tent?

Posted by Mitali Madhusmita 5 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Divya Jatheesh 4 years, 7 months ago

no answer

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Math ch.12 , exer. 12.2 question 3

Posted by Rahul Gupta 5 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Tejaswini Chandok 5 years, 3 months ago

Yr woh toh mention kar diya ho tha tune

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Tejaswini Chandok 5 years, 3 months ago

Which class

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If cos(a+b)=0 show that sin(a-b)=?

Posted by Pradeep Yadav 5 years, 3 months ago

CBSE > Class 10 > Mathematics

3 answers

Gargi Nawghade 5 years, 3 months ago

Cos(a+b)=0 We know that, Cos 90 = 0 So, Cos(a+b) = Cos 90 => a+b = 90 => a = 90 - b Now, Sin(a-b) = Sin(90-b-b) = Sin(90-2b) = Cos 2b [ as Sin(90-x) = Cos x ] Thus, Sin(a-b) = Cos 2b

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Rajesh Sharma 5 years, 3 months ago

cos (a+b) =0 => cos (a+b) = cos 90 => a+b = 90 => a = 90- b Now sin(a+b) = sin(90-b-b) = sin(90-2b) = cos 2b [as sin (90 - x) = cos x]

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Yogita Ingle 5 years, 3 months ago

cos (a+b) =0

=> cos (a+b) = cos 90

=> a+b = 90

=> a = 90- b

Now

sin(a+b)

= sin(90-b-b)