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This is not possible because it is not phythagorian triplet Means sq.of small two digits is not equal to yhe sq.of large digit. i.e Sq.of2+sq.of3=13 Sq.of4=16 Which is not equal.

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In pallagramPQRS l is mid point ofQR prove mis point of trisection on pr

Posted by Vaibhav Saini 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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If sec p + tan p =y Find the value of cosec p

Posted by Teesha Sharma 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Rishabh Jain 6 years, 11 months ago

The value of cosec p = y² + 1/y² - 1.

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If the ratio of the sum of the first n terms of two ap is 7 and + 1 ratio 4 and + 27 then find the ratio of their 9th term

Posted by Aryan Pradhan 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let a₁ and a₂ be the first terms and d₁ and d₂ be the common difference of the two APs respectively.
Let S_n and S'_n be the sums of the first n terms of the two APs and T_n and T'_n be their nth terms respectively.
Then, {tex}\frac { S _ { n } } { S _ { n } ^ { \prime } } = \frac { 7 n + 1 } { 4 n + 27 } \Rightarrow \frac { \frac { n } { 2 } \left[ 2 a _ { 1 } + ( n - 1 ) d _ { 1 } \right] } { \frac { n } { 2 } \left[ 2 a _ { 2 } + ( n - 1 ) d _ { 2 } \right] } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
{tex}\Rightarrow \frac { 2 a _ { 1 } + ( n - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( n - 1 ) d _ { 2 } } = \frac { 7 n + 1 } { 4 n + 27 }{/tex} ........(i)
To find the ratio of mth terms, we replace n by (2m -1) in the above expression.
Replacing n by (2 {tex}\times{/tex} 9 -1), i.e., 17 on both sides in (i), we get
{tex}\frac { 2 a _ { 1 } + ( 17 - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( 17 - 1 ) d _ { 2 } } = \frac { 7 \times 17 + 1 } { 4 \times 17 + 27 } \Rightarrow \frac { 2 a _ { 1 } + 16 d _ { 1 } } { 2 a _ { 2 } + 16 d _ { 2 } } = \frac { 120 } { 95 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + 8 d _ { 1 } } { a _ { 2 } + 8 d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { T _ { 9 } } { T _ { 9 } ^ { \prime } } = \frac { 24 } { 19 }{/tex}
{tex}\therefore{/tex} required ratio = 24:19.

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Give reason.Why maths is boring.mental subject

Posted by Ansh Nanesha 6 years, 11 months ago

CBSE > Class 10 > Mathematics

6 answers

Abhishek Sahu 6 years, 11 months ago

yes, we understand all maths but we have to do more and more practice.

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Amisha Sharma ? 6 years, 11 months ago

Anyone online

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Amisha Sharma ? 6 years, 11 months ago

Same here i too like maths

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@Kumu...#Vk... ?? 6 years, 11 months ago

No maths is not boring..... It is an interesting subject...... I like maths...... ??

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Riya ? 6 years, 11 months ago

Because it refresh and do exercise of our brain in a boring manner.

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Rishabh Jain 6 years, 11 months ago

Maths not a boring subject but it is hard when we do not practice

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How to find root of any number

Posted by Deepak Deepak 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Ansh Nanesha 6 years, 11 months ago

By adding you asw

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Gayatri Gaikwad 6 years, 11 months ago

By prime factorisation method

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Prove that if x and y are odd positive integers, then x2 +y2 is even but not divisible by 4

Posted by Aarushi Yadav 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Aarushi Yadav 6 years, 11 months ago

Thanks

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Ram Kushwah 6 years, 11 months ago

Let the two odd positive no. be x = 2k + 1 and y = 2p + 1

Hence, x² + y² = (2k + 1)² +(2p + 1)²

= 4k² + 4k + 1 + 4p² + 4p + 1

= 4k² + 4p² + 4k + 4p + 2

= 4 (k² + p² + k + p) + 2

clearly, notice that the sum of square is even the no. is not divisible by 4

hence, if x and y are odd positive integer, then x² + y² is even but not divisible by four

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Ansh Nanesha 6 years, 11 months ago

Shut up re

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Find the value of p for which the Equation px2-5x+p=0 has equal roots

Posted by Aman Singh Yadav 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Ram Kushwah 6 years, 11 months ago

Roots of px²-5x+p=0 are equal

if (-5)²=4p(p)

4p²=25

p²=25/4

p=5/2

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Rishabh Jain 6 years, 11 months ago

The value of p is 5/2.

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Is number '7*11*13+13+13*2' a composite number?

Posted by Ayush Sharma 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Yashomani Maurya 6 years, 11 months ago

Yes it is. A composite number is a one which has at least 1 factor other than 1 and itself. Since the given number,i.e.,13(7x11+1+2) is divisible by 13,hence it is a composite number(by definition). Hope it help you. #THANKS!

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Riya ? 6 years, 11 months ago

Yes it is a composite number because it is product of primes. As 7×11×13+13+13×2=13(7×11×1+1+1×2) =13(77+1+2)=13(80)=13×2×2×2×2×5

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How can we solve sin*2 + cos*2 = 1

Posted by Ashish Dubey 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Ram Kushwah 6 years, 11 months ago

by pythagoras theorem

base² + heigh²=hypo²

divide by hypo²

base²/hypo² + height²/hypo²=hypo²/hypo²

sin²x+cos²x=1

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Raushan Kumar 6 years, 11 months ago

By relate trignometery with Pythagoras

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Two no are 18and 24 if hcf is 6find their LCM

Posted by Kavita Khushwaha 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Ram Kushwah 6 years, 11 months ago

LCM*HCF= A*B

SO LCM=18*24/6=72

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Rishabh Jain 6 years, 11 months ago

LCM is equal to the 72

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Draw a triangle ABC with side 6cm,8cm and 9cm and then construct a triangle similar triangle ABC who side are 3/5 corresponding side ABC

Posted by Chetan Sharma 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 11 months ago

IF NEW TRIANGLE IS PQR THEN SIDES ARE

6*3/5,8*3/5,9*3/5

OR 3.6 cm,4.8 cm and 5.4 cm so make the new triangle with these sides

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Prove that in a right angled triangle square of the hyoptenuse is equal to sum of the square of the other two side

Posted by Chetan Sharma 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Prove that the tangents drawn at the ends of diameter of a circle are parallel

Posted by Rishabh Jain 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given: PQ is a diameter of a circle with centre O.
The lines AB and CD are the tangents at P and Q respectively.
To Prove: AB {tex}\parallel{/tex} CD
Proof: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle{/tex}OPA = 90^o ........ (i)
[The tangent at any point of a circle is {tex}\perp{/tex} to the radius through the point of contact]
{tex}\because{/tex} CD is a tangent to the circle at Q and OQ is the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle{/tex}OQD = 90^o ........ (ii)
[The tangent at any point of a circle is {tex}\perp{/tex} to the radius through the point of contact]
From eq. (i) and (ii), {tex}\angle{/tex}OPA = {tex}\angle{/tex}OQD
But these form a pair of equal alternate angles also,
{tex}\therefore{/tex} AB {tex}\parallel{/tex} CD

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Cosis

Posted by Nikhil Raj 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Ayushi Singh 6 years, 11 months ago

Question?

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Cutie Pie ? 6 years, 11 months ago

What's your question Nikhil

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Puja Sahoo 6 years, 11 months ago

Please check the question....

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7th time of 7th term is equal to the 11 time of 11th term find the 18th term

Posted by Sourabh Bhardwaj 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Cutie Pie ? 6 years, 11 months ago

The 18th term will be equal to 0, Sourabh

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Rishabh Jain 6 years, 11 months ago

Because 7a+42d = 11a+110d 4a+68d = 0 a+17d = 0. ( 18 th term = a+17d

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Rishabh Jain 6 years, 11 months ago

18th term is equal to 0

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Show that every positive odd integer in of the form (6m+1) or (6m+3) or (6m+5).

Posted by Pramod Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
{tex}n = 6m + r{/tex}, where {tex}0 \leq r < 6{/tex}
{tex}\Rightarrow{/tex} {tex}n = 6m + r{/tex}, where {tex}r = 0, 1, 2, 3, 4, 5{/tex}
{tex}\Rightarrow{/tex} {tex}n = 6m\ or\ (6m + 1){/tex} {tex}or\ (6m + 2)\ or\ (6m + 3)\ or\ (6m + 4)\ or\ (6m + 5).{/tex}
But, {tex}n = 6m, (6m + 2), (6m + 4){/tex} give even values of n.
Thus, when n is odd, it is of the form {tex}(6m + 1)\ or\ (6m + 3)\ or\ (6m + 5){/tex} for some integer m.

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Probability chapter solution

Posted by Aditya Gupta 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Drashti Vaish 6 years, 11 months ago

You can see the solutions by this app only..

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Formulas of chapter polynomial and linear equation

Posted by Ranbir Yaday Yadav 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Mutyala Gayathri 6 years, 11 months ago

I have that formulas pic how can i send u

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Pls tell chapter vise waitage for board exam of class 10 maths

Posted by Gundeep Singh 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Drashti Vaish 6 years, 11 months ago

You can see by this app only

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Puja Sahoo 6 years, 11 months ago

Number system - 06, Algebra - 20, Coordinate geometry - 06, Geometry - 15, Trigonometry - 12, Mensuration - 10, Statistics and probability - 11 hope it will help you ☺☺☺☺☺

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If sec theta +tan theta=p then find value of cosec theta.

Posted by Harshita Sharma 6 years, 11 months ago

CBSE > Class 10 > Mathematics