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Mean=sum of all the observation/Total no. Of observation. {Median=(n+1/2)" when the observation is odd" and" when even" (n/2+) ,(n/2).} { Mode= an observation with maximum frequency is mode}

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Taran Kamboj 7 years, 8 months ago

Mode=3 median - 2 mean

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If the area of triangle us 0 square units then the vertics of a triangle are

Posted by Karan Garg 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Keshav Bindal 7 years, 8 months ago

Colinear

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LCM of 6,72,and120

Posted by Yashika Solanki 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Shivani K 7 years, 8 months ago

2×2×2×3×3×5=360

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if (ax)2 +bx +c =0 has equal roots . what is the value of c

Posted by Aniket Goyal 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Ashutosh Panda 7 years, 8 months ago

C=b^2/4a

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when will 11th class start

Posted by Aman Ansari 7 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Naaaaaaaaaaaaaaaaaaa Naaaaaaaaaaaaaaaaaaa 7 years, 8 months ago

Around April 15

2Thank You

Zoya Khan 7 years, 8 months ago

I think from June

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find the centre of the circle passing through the points (6,-6)(3,-7)(3,3)

Posted by Sid Mishra 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

Let A → (6, –6), B → (3, –7) and C → (3, 3).
Let the centre of the circle be I(x, y)
Then, IA = IB = IC [By definition of a circle]
{tex}\Rightarrow{/tex} IA² = IB² = IC²
{tex}\Rightarrow{/tex} (x - 6)² + (y + 6)² = (x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²
Taking first two, we get
(x - 6)² + (y + 6)² = (x - 3)² + (y + 7)²
{tex}\Rightarrow{/tex} x² - 12x + 36 + y² + 12y + 36 = x² - 6x + 9 + y² + 14y + 49
{tex}\Rightarrow{/tex} 6x + 2y = 14
{tex}\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]
Taking last two, we get
(x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²
{tex}\Rightarrow{/tex} (y + 7)² = (y - 3)²
{tex}\Rightarrow{/tex} (y + 7) = {tex}\pm{/tex}(y-3)
taking +e sign, we get
y + 7 = y - 3
{tex}\Rightarrow{/tex} 7 = -3
which is impossible
Taking -ve sign, we get
y + 7 = -(y - 3)
{tex}\Rightarrow{/tex} y + 7 = -y + 3
{tex}\Rightarrow{/tex} 2y = -4
{tex}\Rightarrow y = \frac{{ - 4}}{2} = - 2{/tex}
Putting y = -2 in equation (1), we get
{tex}\Rightarrow{/tex} 3x - 2 = 7
{tex}\Rightarrow{/tex} 3x = 9
{tex}\Rightarrow{/tex} x = 3
Thus, I {tex}\rightarrow{/tex} (3, -2)
Hence, the centre of the circle is (3, -2).

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Construct a triangle sides 4,5,6cm and then a triangle similar to it whose sides are2/3 of the corresponding sides of the first triangle

Posted by Sanyam Joshi 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

To construct: To construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are {tex}\frac{2}{3}{/tex} of the corresponding sides of the first triangle.

Steps of construction :

Draw a triangle ABC of sides 4 cm, 5 cm and 6 cm.
From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
Locate 3 points B₁, B₂ and B₃ on BX such that BB₁ = B₁ B₂ = B₂ B₃ .
Join B₃C and draw a line through the point B₂, draw a line parallel to B₃ C intersecting BC at the point C'.
Draw a line through C' parallel to the line CA to intersect BA at A'.
Then, A'BC' is the required triangle.

Justification :
{tex}\because {B_3}C||{B_2}C'{/tex} [By construction]
{tex}\therefore \frac{{B{B_2}}}{{{B_2}{B_3}}} = \frac{{BC'}}{{C'C}}{/tex} [By Basic Proportionality Theorem]
But {tex} \frac{{B{B_2}}}{{{B_2}{B_3}}} = \frac{2}{1}{/tex} [By construction]
Therefore, {tex}\frac{{BC'}}{{C'C}} = \frac{2}{1} \Rightarrow \frac{{C'C}}{{BC'}} = \frac{1}{2} \Rightarrow \frac{{C'C}}{{BC'}} + 1 = \frac{1}{2} + 1{/tex}
{tex} \Rightarrow \frac{{C'C + BC'}}{{BC'}} = \frac{{1 + 2}}{2} \Rightarrow \frac{{BC}}{{BC'}} = \frac{3}{2} \Rightarrow \frac{{BC'}}{{BC}} = \frac{2}{3}{/tex}}...}...} ......(i)
{tex}\because CA||C'A'{/tex} [By construction]
{tex}\therefore \triangle BC'A' \sim \triangle BCA{/tex} [AA similarity]
{tex}\therefore \frac{{AB'}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}} = \frac{2}{3}{/tex} [From eq. (i)]

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What is dharacharya sutra?

Posted by Anish Singh 7 years, 8 months ago

CBSE > Class 10 > Mathematics

6 answers

Annu Chhikara 7 years, 8 months ago

Quadratic formula is -b+-√d/2a

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Anish Singh 7 years, 8 months ago

But it's about general knowledge

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Anish Singh 7 years, 8 months ago

Yes I know

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Annu Chhikara 7 years, 8 months ago

Oh sorry

1Thank You

Anish Singh 7 years, 8 months ago

It means quadric formula

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Annu Chhikara 7 years, 8 months ago

Excuse me wake up ab math ka xam hai

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ANSWER

If two zeroes of a cubic polynomial px3-qx2+rx +s are 2 and -2,then find the third zero of the polynomial in terms p and q

Posted by Vivek Kumar 7 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Maths sabka kitna padh ke hua

Posted by Aurthur Strange 7 years, 8 months ago

CBSE > Class 10 > Mathematics

4 answers

Anupama Chandran 7 years, 8 months ago

8 chap

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Taran Kamboj 7 years, 8 months ago

3 chapter

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Annu Chhikara 7 years, 8 months ago

Same pinch

2Thank You

Kumar Vishal Behera 7 years, 8 months ago

5 chap

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Prove that 1/√2 is a irrational number.

Posted by Sandy Prdh 7 years, 8 months ago

CBSE > Class 10 > Mathematics

3 answers

Pratham Jain 7 years, 8 months ago

It is so much easy

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Pratham Jain 7 years, 8 months ago

Don't worry about it

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Anish Singh 7 years, 8 months ago

Let us prove to the contrary that √3 is irrational u can prove by this process easily

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If d is the HCF of 30,72 find the value of x and y satisfying d=30x+72y

Posted by Ananya Atri 7 years, 8 months ago

CBSE > Class 10 > Mathematics

5 answers

Kumar Vishal Behera 7 years, 8 months ago

But NYRA questn toh aapne galat likha hai

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Kumar Vishal Behera 7 years, 8 months ago

Actually woh lcm hoga naa ki hcf Answer hai x=0,y=5

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Kapil Upadhyay 7 years, 8 months ago

HCF of 45 and 27 45 = (27 x 1) + 18 27 = (18 x 1) + 9 18 = (9 x 2) + 0 Therefore, HCF = 9 9 = 27 – (18 x 1) = 27 – [45 – (27 x 1)] x 1 [since 18 = 45 – (27 x 1)] = 27 – [45 x 1 – 27 x 1 x 1] = 27 – (45 x 1) + (27 x 1) = 27 + (27 x 1) – (45 x 1) = 27{1 + 1} – (45 x 1) = (27 x 2) – (45 x 1) HCF of 45 and 27 in the form of 27x + 45 is (27 x 2) + (45 x -1). Hence, the values of x and y are 2 and -1.Yeh tha iska ans

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Kumar Vishal Behera 7 years, 8 months ago

let me try guys

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Kapil Upadhyay 7 years, 8 months ago

Yahi question maine bhi pucha tha par kisi ne ans nhi diya tha

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ANSWER

For any positive integer n ,prove that n3 -n is divisible by 6

Posted by Pratibha Yadav 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

n³- n = n (n²- 1) = n (n - 1) (n + 1)
Whenever 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, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then 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 + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and 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 + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴ n (n-1) (n+1) = n³- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

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ANSWER

Define Pythagoras theorem by yourself

Posted by Anish Singh 7 years, 8 months ago

CBSE > Class 10 > Mathematics

10 answers

Shivani K 7 years, 8 months ago

a²=b²+c². a=√b²+c²

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Anish Singh 7 years, 8 months ago

Yes bhi sahi hai but there are two ways to prove it one is very easy by method of a+b)12

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Kapil Upadhyay 7 years, 8 months ago

Itni badi theorem yaha kahan se type kru

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Anish Singh 7 years, 8 months ago

Can u prove it

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Shivani K 7 years, 8 months ago

In a right angle triangle the square of the long side (hypotenuse) is equal to the sum of the square two other sides. " a²=b²+c²"

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Kapil Upadhyay 7 years, 8 months ago

Yes

1Thank You

Anish Singh 7 years, 8 months ago

Have u answered by yourself

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Anish Singh 7 years, 8 months ago

You are correct

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Kapil Upadhyay 7 years, 8 months ago

Square of hypotenuse is equal to sum of square of other two sides

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Anish Singh 7 years, 8 months ago

The square of hypotenuse is equal to the sum of square of rest two sides

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ANSWER

Blue print in maths

Posted by Devesh Yadav 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Foram Desai 7 years, 8 months ago

Check the syllabus of maths on the app itself.... U will get it...

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In a quadrilateral ΔBCD, ∠A+ ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2. ................ Ise solve karne ka method bta do koi pls........

Posted by Kapil Upadhyay 7 years, 8 months ago

CBSE > Class 10 > Mathematics

6 answers

Kapil Upadhyay 7 years, 8 months ago

Yeh tumne kaha kya samajh nhi aaya (is chapter much move on kya h yeh)

1Thank You

Aakarsh Patel 7 years, 8 months ago

Ulta kr lo niche se

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Anish Singh 7 years, 8 months ago

Can u answer my questions

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Anish Singh 7 years, 8 months ago

Is chapter see much nahi hoga move on

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Kapil Upadhyay 7 years, 8 months ago

Lagta h kisi ko nhi aata.........??????????????

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Kapil Upadhyay 7 years, 8 months ago

Koi batayega bhi.................???????????

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Two tangent TP andTQ are drawn to a circle with center O from an external point T prove that angle PTQ=2OPQ

Posted by Amit Mall 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Anupama Chandran 7 years, 8 months ago

Given: TP and TQ are tangents To prove: PTQ =2OPQ Proof: in ∆TPQ TP=TQ =>angleP=angleQ Let <P=<Q=<i(any angle) <PTQ=180-2¡ OP is the radius and TP is the tangent <OPT=90° <OPQ=90°-¡ 2<OPQ=180-2¡ From this PTQ=2OPQ

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ANSWER

Find the roots of the quadratic equation x square + (a+b ÷ a+a÷a+b)x +1=0

Posted by Garima Gupta 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

{tex}x ^ { 2 } + \left( \frac { a } { a + b } + \frac { a + b } { a } \right) x + 1 = 0{/tex}

{tex}x ^ { 2 } + \frac { a } { a + b } x + \frac { a + b } { a } x + 1 = 0{/tex}
{tex}x \left( x + \frac { a } { a + b } \right) + \frac { a + b } { a } \left( x + \frac { a } { a + b } \right) = 0{/tex}
{tex}\left( x + \frac { a } { a + b } \right) \left( x + \frac { a + b } { a } \right) = 0{/tex}
{tex}\therefore{/tex} {tex}x = \frac { - a } { a + b } , \frac { - ( a + b ) } { a }{/tex}

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In the given figure OB is perpendicular bisector of the line segmentDE,FA is perpendicular too but and FE intersect OB at c. Prove that 1/OA+1/OB=2/OC

Posted by Bharath Challa 7 years, 8 months ago

CBSE > Class 10 > Mathematics