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Yogita Ingle 6 years, 10 months ago

Given that the sum of two numbers is 9 Let the two numbers be x and 9-x
By the given hypothesis, we have
1/x + 1/9-x = 1/2
(9 - x + x)/x(9-x) = 1/2
18 = 9x - x²
x² - 9x + 18 = 0
(x - 6)(x - 3) = 0
x = 6 or x = 3
Therefore the numbers are 3 and 6.

ANSWER

If the points p(-3,9),q(a,b)and R(4,-5)are collinear and a+b=2 find the value of A and B

Posted by Anshika Saini 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

I guess the correct question is :

Q: If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a+b=1, find the value of a and b.

ANSWER

Find HCF of 65 and 117 and find a pair of integral values of m&n such that HCF =65m+117n.

Posted by Nikita Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

<font face="Arial, sans-serif"><font style="box-sizing: inherit; outline: none; user-select: initial !important; max-width: 100%; overflow: hidden;">By Euclid's division algorithm

117 = 65x1 + 52.</font></font>

65 = 52x1 + 13

52 = 13x4 + 0

Therefore 13 is the HCF (65, 117).

Now work backwards:

13 = 65 + 52x(-1)

13 = 65 + [117 + 65x(-1)]x(-1)

13 = 65x(2) + 117x(-1).

∴ m = 2 and n = -1.

ANSWER

Find HCF of 378,180, 420. Is HCF *LCM is equal to the product of three numbers?

Posted by Nikita Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Prashant Parihar 6 years, 10 months ago

Factor all three numbers common number in all Three number then hcf and power mutiply then lcm Lcm×hcf=a×b×c

Gaurav Seth 6 years, 10 months ago

378 = 2x3x3x3x7

180 = 2x2x3x5x7

420 = 2x2x3x5x7.

L.C.M ( 378,180,420) = 2x2x3x3x3x5x7 = 3780

H.C.F ( 378,180,420) = 2x3 = 6

⇒ L.C.M ( 378,180,420) x H.C.F ( 378,180,420) ≠ 378 x 180 x 420.

⇒ 3780 x 6 ≠ 378 x 180 x 420.

∴ 22680 ≠ 28576800.

∴HCF and LCM of three number is not equal to product of three number.

ANSWER

Determine the values of y and q so that the prime factorization of 2520 is expressible as 23*y*q*7.

Posted by Nikita Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Nikita Sharma 6 years, 10 months ago

But 23 is not its factor

Gaurav Seth 6 years, 10 months ago

Since, 2520 = 2 x 2 x 2 x 3 x 3 x 5 x 7
= 23 x 32 x 5 x 7

Therefore, p = 2 and q = 5

ANSWER

If secθ + tanθ = p, then find the value of cosecθ ??

Posted by Yash Goyal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 10 months ago

Given, sec θ + tan θ = p ...........1

We know that

sec² θ - tan² θ = 1

=> (sec θ - tan θ)(sec θ + tan θ) = 1

=> (sec θ - tan θ) * p = 1

=> sec θ - tan θ = 1/p ............2

Add equation 1 and 2, we get

2 * sec θ = p + 1/p

=> 2 * sec θ = (p² + 1)/p

=> sec θ = (p² + 1)/2p

and cos θ = 1/sec θ

=> cos θ = 2p/(p² + 1)

Subtract equation 1 and 2, we get

2 * tan θ = p - 1/p

=> 2 * tan θ = (p² - 1)/p

=> tan θ = (p² - 1)/2p

and cot θ = 1/tan θ

=> cot θ = 2p/(p² - 1)

We know that

tan θ = sin θ/ cos θ

=> sin θ = tan θ * cos θ

=> sin θ = {(p² - 1)/2p} * {2p/(p² + 1)}

=> sin θ = (p² - 1)/(p² + 1)

and cosec θ = 1/sin θ

=> cosec θ = (p² + 1)/(p² - 1)

ANSWER

Prove that the product of any 3 consecutive positive integer is divisible by 6

Posted by Nikita Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 10 months ago

Let three consecutive positive integers be, n, n + 1 and n + 2.
When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
Similarly, when a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
Hence n (n + 1) (n + 2) is divisible by 2 and 3.
∴ n (n + 1) (n + 2) is divisible by 6.

ANSWER

Find the value of : (-1) + (-1)2n +(-1)2n+1 +(-1)4n+1 where n is positive odd integer.

Posted by Nikita Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Nikita Sharma 6 years, 10 months ago

Isme 2n, 2n+1,4n+1 inke exponents hai

ANSWER

Find the root of the equation 1/x-3 + 1/x-5=1 ,x is not equal to 3/2 ,5

Posted by Harshita Jaiswal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Areeba Siddiqui 6 years, 10 months ago

1/x-3 +1/x-5=1 (x-5)+(x-3)/(x-3)(x-5)=1 X-5+x-3=xroot-8x+15 Xroot-10x+23 (by b root-4ac,-b+-root D) 5+root2,5-root2answers

ANSWER

A Sector is cut from a circle of radius 21 cm. The angle of the sector is 150' . Find the length of the arc and the area of the sector

Posted by Simple Simplee 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Ram Kushwah 6 years, 10 months ago

Arc =150/360×2×22/7×21 cm

=17.5 cm

Area =150/360*πr^2

=150/360×22/7*×21×21

=577.5 cm sq

Chetna Pandey 6 years, 10 months ago

Use formula ø/360πrsquare for finding the area of sector

ANSWER

In equilateral triangle ABC ,D is a point on side BC such that BD = 1/3 BC .prove that 9AD^2 =7AB^2.

Posted by Aditi Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Arjun Senthil 6 years, 10 months ago

It is a book sum.search ncert solutions and find answers.

ANSWER

2^2x^2-7x+5=1 evaluate

Posted by Sakshi Rathore 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Chetna Pandey 6 years, 10 months ago

x = -4

ANSWER

यदि दो समरूप Tribhuj ke sangat madhiko ka anupat 5:7 ho to inki bhujao ka anupat gat karo

Posted by Aarti Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

4 answers

Aarti Sharma 6 years, 10 months ago

Ansawr

..... ...... 6 years, 10 months ago

5:7 ही होगा। क्योंकि sides और median का अनुपात(ratio) similar triangles में same होता है।

Vivek Kumar 6 years, 10 months ago

i can't know your question

Aarti Sharma 6 years, 10 months ago

ANSWER

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side

Posted by Laxmi Kumari 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Chetna Pandey 6 years, 10 months ago

It's BPT's converse theorem which will not come in exams to proove but this theorem is applicable in chapter ∆ in some questions ..........if u want this theorem just Refer your NCERT....

ANSWER

If sintheta equalto cos theta ,then find the valueof 2tantheta +cossquare theta

Posted by Suraj Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Gungun_ ♥️♥️♥️ 6 years, 10 months ago

5/2 will b d ryt answer

Chetna Pandey 6 years, 10 months ago

5/2 is correct answer

Chetna Pandey 6 years, 10 months ago

5/2 in place of theetha put 45 because cos = sin at angle 45°

ANSWER

Find all zeroes of of the polynomial (2x4-9x3+5x2+3x-1) if two zeroes are (2+

Posted by Vaishnavi Jadhav 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given:
f(x) = (2x⁴ – 9x³ + 5x² + 3x – 1)
Zeroes = (2 + √3) and (2 – √3)
Given the zeroes, we can write the factors = (x – 2 + √3) and (x – 2 – √3)
{Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}
Multiplying these two factors, we can get another factor which is:
((x – 2) + √3)((x – 2) – √3) = (x – 2)^{2 –} (√3)²
⇒x² + 4 – 4x – 3 = x² – 4x + 1
So, dividing f(x) with (x² – 4x + 1)

f(x) = (x² – 4x + 1) (2x² – x – 1)
Solving (2x² – x – 1), we get the two remaining roots as

{tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}
where f(x) = ax² + bx + c = 0(using Quadratic Formula)

{tex}\mathrm{x}=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(2)(-1)}}{2(2)}{/tex}
{tex}\mathrm{x}=\frac{-1 \pm 3}{4}{/tex}
{tex}\Rightarrow \mathrm{x}=1,-\frac{1}{2}{/tex}
Zeros of the polynomial = {tex}1,-\frac{1}{2}, 2+\sqrt{3}, 2-\sqrt{3}{/tex}

ANSWER

If cos(a+b)=0, show that sin (a-b)=cos 2b

Posted by Priyanshu Golu 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Sonu Kumar 6 years, 10 months ago

Cos(a+b)=0 Cosa+cosb=0 Cosa= -cosb We take rhs Sin(a-b) Sina-sinb Cos(90-a) -cos(90-b) Cos90-cosa -cos90+cosb -cosa+cosb -(-cosb) + cosb Cosb+ cosb Cos(b+b) Cos2b prooved

Avinash Saigal 6 years, 10 months ago

Cos(a+b)= 0 (a+b)= 90 a=(90-b) Sin(a-b)= sin ( 90-2b) Since, sin (A-B)= sinA* cosB - cosA* sinB Sin(90-2b)= sin90*cos2b - cos90 -sin2b 1*cos2b -0*sin2b Hence answer is cos2b...... Thank you??

ANSWER

AD is an altitude of an equilateral triangle ABC. On AD as base,another equilateral triangle ADE is constructed. Prove that Area(ADE):Area(ABC) =3:4

Posted by Muskan Muskan 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mayank Mohit Agarwal 6 years, 10 months ago

In tri.ABD <abd=60• So, tan 60•=AD/BD √3=AD/BD √3BD=AD 2√3BD=2AD √3BC=2AD AD/BC=√3/2 Now,ar(ADE)/ar(ABC)=AD^2/BD^2 So, ar(ADE)/ar(ABC)=(√3/2)^2 =3/4 So, the ratio is 3:4.

ANSWER

Y= -1cube -4 multiply1

Posted by Abhishek Maurya 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find The HCF Of K, 2k,3k,and 5k, where k is a positive integer

Posted by Milan Deshwal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

The HCF of k,2k,3k,4k and 5k will be k as the following no.s are consecutive and only k is a common factor as all numbers are multiplied by it.

ANSWER

How can prove? two similar triangles have also congruent triangle

Posted by Prashant Parihar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Purvanshi Yadav 6 years, 10 months ago

By SSS(Side side side) or AAA(Angle angle angle) criteria

ANSWER

What are included in chapter Real numbers

Posted by Shrijith The Artist ??? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Shrijith The Artist ??? 6 years, 10 months ago

Mtlb creal numbers chapter me kya he (formula, sign's, etc)

ANSWER

*Q. The sum of first q terms of an AP is 162. The ratio of its 6th term to its 13th term is 1:2. Find the first and 15th term of the AP.* (CBSE 2015) *ANS. 6,48*

Posted by Karan Kumar Mahato 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

if an isosceles triangle ABC in which AB=AC=6cm , is inscribed in a circle of radius 9 cm , find the area of triangle.

Posted by Prajkt Sonkusare 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Avinash Saigal 6 years, 10 months ago

Ans. Given : A isoscele triangle ABC in which AB = AC = 6cm is inscribed in circle with center O and radius 9 cm. let AM = x cm In Quadrilaterla OABC OB = OC = 9 cm and AB = AC = 6 cm. it means it is Kite and diagonals of kite are perpendicular to each other, So OA ⊥ BC In Right∆AMC, => AM2 + MC2=AC2 ........ [Pythagoras Theorem] => MC2 = 62 - x2 = 36-x2 ........... (1) in Right∆OMC, => OM2 + (MC)2 = OC2 => => (9-x)2 + 36-x2 = 81 => => 81 + x2 -18x + 36 - x2 = 81 => 18x = 36 => x = 2 AM = 2 cm Now, MC = 62-22 = 36-4 = 32 = 42 cm As OM is Perpendicular to Chord BC So M is midpoint of BC. SO BC = 2 * MC = 2×42 = 82 cm Area of ∆ABC = 12×BC×AM = 12×82×2 = 82cm2 thank you???

Alinarathore Alinarathore 6 years, 10 months ago

ANSWER

99×.

Posted by Tripti Saxena 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Tripti Saxena 6 years, 10 months ago

4Thank You

ANSWER

Solve for x: 1/2a+b+2a=1/2a+1/b+1/2x.

Posted by Khushi Star ? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

1/2a+b+x=1/2a +1/b +1/x correct question, ye hai, ye mere e am me aya tha......, and answer hai x=-2a, -b........

ANSWER

Solve the equation 1/x+1+2/x+2+4/x+4,x is not equal -1,-2,-4?

Posted by Khushi Star ? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

5 answers

Prashant Parihar 6 years, 10 months ago

Solve equation by simply by taking lcm then find only value of x

Khushi Star ? 6 years, 10 months ago

You can solve complete question

Khushi Star ? 6 years, 10 months ago

Thanks

Khushi Star ? 6 years, 10 months ago

Please help

Khushi Star ? 6 years, 10 months ago

Don't know

ANSWER

Prove that sum of squares of diagonals of parallelogram are equal to sum of squares of its sides

Posted by Khushi Bhati 5 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Ansh Bhatnagar 6 years, 10 months ago

Here ,a short ans for this question we have to prove that ab2+bc2+cd2+da2=ac2+bd2 since

Avinash Saigal 6 years, 10 months ago

In parallelogram ABCD, AB = CD, BC = AD Draw perpendiculars from C and D on AB as shown. In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem] ⇒ AC2 = (AB + BE)2 + CE2 ⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2 → (1) From the figure CD = EF (Since CDFE is a rectangle) But CD= AB ⇒ AB = CD = EF Also CE = DF (Distance between two parallel lines) ΔAFD ≅ ΔBEC (RHS congruence rule) ⇒ AF = BE Consider right angled ΔDFB BD2 = BF2 + DF2 [By Pythagoras theorem] = (EF – BE)2 + CE2 [Since DF = CE] = (AB – BE)2 + CE2 [Since EF = AB] ⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2 → (2) Add (1) and (2), we get AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2) = 2AB2 + 2BE2 + 2CE2 AC2 + BD2 = 2AB2 + 2(BE2 + CE2) → (3) From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem] Hence equation (3) becomes, AC2 + BD2 = 2AB2 + 2BC2 = AB2 + AB2 + BC2 + BC2 = AB2 + CD2 + BC2 + AD2 ∴ AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. Thank you?

ANSWER

In triangle ABC it is given that DE || BC . If AD=3 & DB=2 & DE=6 then find BC

Posted by Yogesh Parihar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Neha Sirur 6 years, 10 months ago

By aaa criteria...triamgle ade~abc Ad/ab=de/bc...2/5=6/bc...hence bc=15cm...

Gaurav Seth 6 years, 10 months ago

Given: ABC is a triangle , DE parallel BC and AD= 3 cm, DB=2 cm and DE=6 cm

To find: AE

sol: Since DE parallel BC

angle ADE= angle ABC (corresponding angles)

and angle AED = angle ACB ( " )

Triangle ADE ≈ triangle ABC ( by AA similarity)

therefore , AD/AB=DE/BC=AE/AC (1)

From (1)

AD/AB=AE/AC

2/5=x/18

2×18=5x

36=5x

x=36/5

x=7.5cm

∵ AE = 7.5 cm

Priyanshi Khandelwal 6 years, 10 months ago

ANSWER

If x=a sec A+ b tan A and y=a tan A+ b sec A, prove that x^2-y^2=a^2-b^2

Posted by Aayu... Sad Girl....??? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Purvanshi Yadav 6 years, 10 months ago

X= a secA+b tanA Y= a tanA+b secA On squaring we get the following; X2+Y2= a2 sec2A+b2 tan2A - (a2 tan2A+b2 sec2A) X2+Y2=a2 sec2A+b2 tan2A - a2- tan2A-b2-sec2A X2+Y2= a2-b2 hence proved.