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

The standard form of a quadratic equation is ax²+bx+c=0, where a,b and c are real numbers and a≠0. 'a' is the coefficient of x². It is called the quadratic coefficient.

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A and b are the zero of the quadratic polynomial

Posted by Kanak Parmar Parmar 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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We are given a circle with O, an external point T and two tangents TP and TQ to the circle,where P,Q are the point of contact .We need to prove that

Posted by Ravi Ooviya 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years, 1 month ago

Two tangents TP and TQ are drawn to a circle with center O from an external point T. Prove that angle PTQ = 2 OPQ.

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)

∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)

∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))

⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠OPT = 90º

⇒ ∠OPQ + ∠TPQ = 90º

⇒ ∠OPQ = 90º – ∠TPQ

⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)

From (1) and (2), we get

∠PTQ = 2∠OPQ

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Find the discriminant of the quadratic equation 2x^2+4x+5=0

Posted by Sneha Thapalyal 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Tarannum Jahan 5 years, 1 month ago

2x²+4x+5 =0 Comparing with ax²+bx+c We get a=2 ,b=4,c=5 By discrimimant formula b²-4ac =4²-4×2×5 =16-40 =-24 Ans

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A K47 5 years, 1 month ago

-24 using b^2-4ac

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Find the angular elevation of the sun when the shadow of a 10m long pole is 10√3

Posted by Nilakshi Dutta 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Find zeroes of quadratic polynomial 9tsquare-6t+1 and verify their relationship between zeroes and coefficients

Posted by Abi Naya 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years, 1 month ago

9t² - 6t + 1 = 0

9t² - 3t -3t + 1 = 0

3t(3t -1) -1 (3t-1) =0

(3t-1)(3t-1) =0

Both roots are same and = t = 1/3.

Sum of the roots = -b/a = -(-6/9) = 2/3

1/3 + 1/3 = 2/3

Product of roots = (c/a) = 1/9

1/3 x 1/3 = 1/9.

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first exercise

Posted by Prithpal Singh Sikkh 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years ago

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 225

Solutions:

i. 135 and 225

As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220

In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.

Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 225

As we know, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,

225 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 225 is 51.

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S=12 Upon 2 big bracket 2multiply-7 small bracket 12-1bracket close -70

Posted by Singh Alok 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Paru ? 5 years, 1 month ago

Is your answer is this ??. S=12[2×(-7){12-1}]-70

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State state whether the equation (x + 1)( x - 2) +x=0 has two distinct real roots are not justify your answer

Posted by Kaushlendra Pratap 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years, 1 month ago

Q u e s t i o n : State whether the equation (x+1) (x - 2) + x = 0 has two distinct real roots or not. Justify your answer.

A n s w e r :

So, Given equation has two distinct real roots.

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Prove that (Tan^2A-tan^2B)=(sin^2A-sin^2B)upon (cos^A*cos^2B)=(cos^2B-cos^A) upon (cls^B*cos^A)

Posted by Satyam Kumar 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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If 1+sin^2theta=3sintheta*costheta then prove that tantheta=1 or 1/2

Posted by Satyam Kumar 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years ago

1+sin^2 theta=3 sin theta cos theta (we know that sin^2 theta + cos^2 theta =1)

= ( sin^2 theta + cos^2 theta ) + sin ^2 theta = 3 sin theta cos theta

= sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta

= cos^2 theta + 2 sin^2 theta = 3 sin theta cos theta

On dividing by cos^2 theta, we get

= 1 + 2 tan^2 theta = 3 tan theta

Let tan theta = b

2b^2 - 3b + 1 = 0

= (2b-1)(b-1) = 0

b = 1 or 1/2

So, tan theta = 1 or 1/2.

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2 cubes each of volume 64 cm^3 are joined end to end . Find the surface area of resulting cuboid.

Posted by Harshit Badgujjar 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Harshit Badgujjar 5 years, 1 month ago

ANSWER The volume of cube 64cm3 Side of cube 364=4cm Length of resulting cuboid 4+4=8 Surface area =2(lb+hl+bh) =2(4(4)+4(8)+8(4))=2(16+32+32)=2(80)=160cm2

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Prove that cos a upon 1 + sin a + 1 + sin a upon cos a is equal to 2 sec a

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Swati Roy 5 years, 1 month ago

L. H.S Cos a upon 1 + sin a+ 1 + sin a upon Cos a = cos square A + (1 + sin square a ) upon ( 1 + sin a) Cos a = cos square a + 1+ s i n square a 2 sin a upon ( 1 + sin a) + Cos a = sin square a+ cos square a+ 1 + 2 sin a upon ( 1+ sin a ) + Cos a = 1+1 + 2 sin a upon (1 + sin a) Cos a= 2 + 2 sin a upon (1+sin a) Cos a = 2 ( 1+sin a) + cos a = 2 •1 upon Cos a = 2sec a

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ABC is an isosceles triangle right angled at c prove that a b square is equal to 2 AC square

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Adarsh Patel 5 years, 1 month ago

Copying all vidya mandir mid term exam questions , ?????????

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Swati Roy 5 years, 1 month ago

Given ABC is an Isosceles triangle right angled at c prove a b square is equal to 2 AC square proof (AB)² =(AC)² + (BC)² =(AB)²= (AC)²+ (AC)² [BC=AC] = (AB)²= 2(AC)² proved

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Sristi Tekriwal 5 years, 1 month ago

Copying all vidya mandir mid term exam questions , ?????????

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Explain why 7 *11 * 13 + 13 and 7 * 6 *5 * 4 *3 * 2 *1 + 5 are composite number

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

. . 5 years, 1 month ago

In the first no. 13 is common factor 13(7*11*1+1) In the second one 2 is common factor. 2(7*3*5*2*3*1*1+5) So they are composite numbers

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If tan a is equal to COT b prove that a + b is equal to 90 degree

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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A rhombus of side 10 centimetre one of the diagonal is 12 centimetre long find the length of second

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years ago

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.

Also, diagonals of a rhombus bisect each other at right angles.

If AC = 12 cm, AO = 6 cm

Applying Pythagoras theorem in right-angled triangle AOB. We get:

Hence, the length of the second diagonal BD is 16 cm.

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Find the sum and product of zeros of 3x^2-5x+6

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years ago

we have to find sum and product of zeroes of the given polynomial 3x² - 5x + 6.

we know, for general form of quadratic polynomial, ax² + bx + c

sum of zeroes = - coefficient of x/coefficient of x² = -b/a
sum of zeroes = - coefficient of x/coefficient of x² = -b/a product of zeroes = constant/coefficient of x² = c/a

on comparing 3x² - 5x + 6 with ax² + bx + c we get, a = 3, b = -5 and c = 6

so, sum of zeroes = -b/a = -(-5)/3 = 5/3

product of zeroes = c/a = (6)/3 = 2

hence, sum of zeroes = 5/3

and product of zeroes = 2

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The decimal expansion of the number 43/2^4*1/5^3

Posted by Yadav Ahir. 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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can we follow someone on this app or see someones profile ☺

Posted by Itz Raman Here ☺ 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Payal Dhiman 5 years, 1 month ago

Ooh oh

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Satyam Kumar 5 years, 1 month ago

We can't follow anyone.?

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: (ᵔᴥᵔ) ???? (ᵔᴥᵔ) 5 years, 1 month ago

According to me..?..Nope we can't follow someone and also can't see someones profile.?

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arr koi mainu dsdo mere pehlan wale question da answer dedo koi. je kise nu anda ae?

Posted by Itz Raman Here ☺ 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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hlo guys gud afternoon to all my slf raman .sometimes i am puzzle in some questions that where should i found hcf and where Lcm tell me guys. have a nyc day

Posted by Itz Raman Here ☺ 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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The point A(9,0),B(9,6)C(-9,6)&D(-9,0)are the vertices of A

Posted by Aditi Raj 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Gaurav Seth 5 years, 1 month ago

AB=square root of((9-9)^2+(6-0)^2)=6
BC=square root of((9-(-9))^2+(6-6)^2)
=square root of (18^2+0)=18
CD=square root of((-9-(-9))^2+(6-0)^2)
=square root of(0+6^2)=6
AD=square root of((9-(-9))^2+(0-0)^2)
=square root of(18^2+0)=18
Since, AB=CD AND BC=AD
THEREFORE, A,B,C,D ARE THE VERTICES OF A RECTANGLE.

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the discrimination of the quadratic equation is

Posted by Zakcheng Gabil 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Lakshy Choudhary 5 years, 1 month ago

D=b^2-4ac

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Gaurav Seth 5 years, 1 month ago

The Discriminant Formula in the quadratic equation ax² + bx + c = 0 is

△ = b² − 4ac

If the discriminant value is positive, the quadratic equation has two real and distinct solutions.
If the discriminant value is zero, the quadratic equation has only one solution or two real and equal solutions.
If the discriminant value is negative, the quadratic equation has no real solutions.

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the discrimination of the quadratic equations is

Posted by Zakcheng Gabil 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Charu Gupta 5 years, 1 month ago

B²-4ac

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Gaurav Seth 5 years, 1 month ago

The Discriminant Formula in the quadratic equation ax² + bx + c = 0 is

△ = b² − 4ac

If the discriminant value is positive, the quadratic equation has two real and distinct solutions.
If the discriminant value is zero, the quadratic equation has only one solution or two real and equal solutions.
If the discriminant value is negative, the quadratic equation has no real solutions.

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Hi . Please tell best book for standard maths class 10 . And I am good at studies and want 90+ in boards . RS or rd

Posted by Manya Mahajan 5 years, 1 month ago

CBSE > Class 10 > Mathematics