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

Mane toh board ka paper da diaya ha ab mai 11th mai hu or mane last year diya tha sabhi questions ncert mae sa aye tha mane pass question paper bhi ha

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Miss. 1234 6 years, 1 month ago

Yeah Sahil.....NCERT hi krti hu ...but sst maths or science me ncert book ke alwa b krti hu.... because board paper kahi se b aa skta h??Aisa teacher ka manana h....aur mai teacher ki baat se puri trh sehmtt hu..toh ncert ke alwa b RD in maths, S. Chand in science ,Arihant in Sst ......inn 3 books ko b pdhti hu???ap b inn books ko pdhna bhot achhi h aur Arihant to best h sst ke liye isme apke mostly answers point me mil jayenge jo net pe mil pana mushkil h............Thank u...meri speech yahi smapt hoti h?

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Sahil Singh 6 years, 1 month ago

Exam k liya ncert kar lo ohi kafi ha paper ncert mai se hi aata ha

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Miss. 1234 6 years, 1 month ago

Nhii diya.....apne diya h kya? Konsa test ? Kesa test??? Kuch mujhe b bta do kis type ke ques. hote h isme ncert book se related ya time pass test...?????????????apki ati kripa hogi ......by Miss 1234

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Brainly Buddy 6 years, 1 month ago

Ok sis

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Angel Niddhi 6 years, 1 month ago

ha pata to ahi lekin , questions telle karana hai matlab nahi samjha . Hamare school me aj arayabhatt test ke bare me bataya hai kyo bhai tumhe kisne bataya ? Tumhara question phir ek bar repeat karo woh bhi ache se samajane me aye aise , ok by

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Brainly Buddy 6 years, 1 month ago

Nhi diya?

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Miss. 1234 6 years, 1 month ago

Okay

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Brainly Buddy 6 years, 1 month ago

But hua h

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Brainly Buddy 6 years, 1 month ago

Exact naam yaad ni

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Miss. 1234 6 years, 1 month ago

No.....Which Aryabhata test??

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ANSWER

what is the probalite of getting 3

Posted by Pravesh Sajwan 6 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Prove that the radius is perpendicular to tangent on point of contact

Posted by Gopi Raman 6 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Tursha Fulia 6 years, 1 month ago

Refer maths textbook pg no.-208 theorem 10.1

3Thank You

ANSWER

3 consecutive natural no. Are such that the square of the middle no. Exceed the difference of the square of other 2 by 60 Find the no.

Posted by Gunjan Lathwal 6 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find the ratio in which the line segment joining (-2,-3) and (5,6) is divided by (i) X- axis (ii) Y- axis Also, find the coordinates of the points of division in each case

Posted by Anshul Kumar Parira 6 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 1 month ago

we know that equation of line

y = mx + c

where c is constant

{tex}m = \frac{y2 - y1}{x2 - x1} \\ m = \frac{6 - ( - 3)}{5 - ( - 2)} \\ m = \frac{9}{7}{/tex}

y = (9/7)x + c

when x = 5 then y = 6

6 = (9/7)×5 + c

=> 42 = 45 + 7c

=> c = -3/7

equation is

y = (9/7)x - 3/7

7y = 9x - 3

x axis interception means y = 0

=> 0 = 9x -3

=> x = 3/9 = 1/3

y axis interception means x = 0

=> 7y = -3

=> y = -3/7

y axis intercepted at (0, -3/7)

Distance from (-2,-3)

{tex}= \sqrt{(-2-0)^{2} + (-3 -(-3/7)^2} = \sqrt{(-2)^2 + (\frac{-18}{7})^2 } \\= \sqrt{4 + \frac{324}{49} } = \sqrt{\frac{520}{49}}{/tex}

Distance from (5, 6)

{tex}= \sqrt{(5-0)^{2} + (6 -(-3/7)^2} = \sqrt{(5)^2 + (\frac{45}{7})^2 } \\= \sqrt{25 + \frac{2025}{49} } = \sqrt{\frac{3250}{49}}{/tex}

{tex}Ratio = \sqrt{\frac{3250}{49}} / \sqrt{\frac{520}{49}} = \sqrt{6.25} = 2.5{/tex}

x axis interception at (1/3 , 0)

{tex}Distance from (-2,-3) = \sqrt{(-2- \frac{1}{3})^2 + (-3-0)^2} = \sqrt{\frac{49}{9} + 9} = \sqrt{\frac{130}{9}}{/tex}

{tex}Distance from (5,6) = \sqrt{(5- \frac{1}{3})^2 + (6-0)^2} = \sqrt{\frac{196}{9} + 36} = \sqrt{\frac{520}{9}}{/tex}

{tex}Ratio = \sqrt{\frac{520}{130}} = \sqrt{4} = 2{/tex}

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ANSWER

S3=3(S2-S1)

Posted by Himanshi Thakkar 6 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 1 month ago

Q : The sum of n, 2n ,3n terms of an AP are S1,S2,S3,respectively. Prove that S3 =3(S2- S1)

Answer:

Let ‘a’ be the first term of the AP and ‘d’ be the common difference

S₁ = (n/2)[2a + (n – 1)d] --- (1)

S₂ = (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2)

S₃ = (3n/2)[2a + (3n – 1)d] --- (3)

Consider the RHS: 3(S₂ – S₁)