Ask questions which are clear, concise and easy to understand.

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ANSWER

Posted by Rishvanth Rajaa an hour ago

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Priyan Villain an hour ago

Anyone can learn maths whether they're in higher maths at school or just looking to brush up on the basics.After discussing ways to be a good math student ,this article will teach you the basic progression of maths courses and will give you the basic elements that you'll need to learn in each course.then the article will go through the basics of learning arthimetic which will help both kids and elementary school and anyone else who needs to brush up on the fundamentals.

Posted by Parag Jhariya 14 hours ago

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Posted by Akash Yadav 1 day, 3 hours ago

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Jagtar Mistri 1 day, 2 hours ago

Mathematical induction is a mathematical proof technique. It is used to prove things.

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Sia 🤖 1 day, 4 hours ago

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time.

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Posted by Anushka Shekhawat 2 days, 7 hours ago

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Sia 🤖 2 days, 7 hours ago

1, A_{1}, A_{2}, A_{3}, ----- A_{m}, 31 are in AP.

a = 1

a_{n} = 31

{tex}a _ { m + 2 } = 314{/tex}

a_{n} = a + (n - 1)d

31 = a + (m + 2 - 1)d

{tex}d = \frac { 30 } { m + 1 }{/tex}

{tex}\frac { A 7 } { A _ { n - 1 } } = \frac { 5 } { 9 }{/tex} (Given)

{tex}\frac { 1 + 7 \left( \frac { 30 } { m + 1 } \right) } { 1 + ( m - 1 ) \left( \frac { 30 } { m + 1 } \right) } = \frac { 5 } { 9 }{/tex}

m = 1

Posted by Jitendra Kumar 2 days, 16 hours ago

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Posted by Jyoti Gupta 3 days, 7 hours ago

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Posted by Shilpa Kumari 3 days, 12 hours ago

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Gaurav Seth 3 days, 7 hours ago

uppose,

A={1,2,3,4,5,6}

B={2,4,6}

In this condition, every element in B is contained in A so,

B(A

But however, B is the set of even numbers and A is the set of natural numbers and thus they are very different in terms of n.

We write even number as 2n whereas natural number just n.

So, B doesn't belong to A.

Rahul Raj 3 days, 12 hours ago

1. "Belongs to".....if we say a belongs to A , it means a is itself an element of A or a is a member of A. 2."subset"......it is defined as part of a set.If we say set B is a subset of set A it means each element of B is a member of set A. Btt meanings r same nearly btt what is different in both......ex- if A ={1,2,3,4} then u can simply write 2 belongs to A btt to write as a subset u will have to write { 2} is a subset of A. ......well it is a nice question.it made think a little longer😊😊 hope u will understand...

Posted by Birendar Kumar 3 days, 22 hours ago

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Gaurav Seth 3 days, 14 hours ago

Any of the ways we can arrange things, where the order is important.

Example: You want to visit the homes of three friends ("a"), ("b") and ("c"), but haven't decided in what order. What choices do you have?

Answer: {a,b,c} {a,c,b} {b,a,c} {b,c,a} {c,a,b} {c,b,a}

Posted by Isha Suryan 3 days, 11 hours ago

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Sia 🤖 3 days, 11 hours ago

Let P(n) = a + ar + ar^{2} + ..... + ar^{n-1}{tex} = \frac{{a({r^n} - 1)}}{{r - 1}}{/tex}.

For n = 1

{tex}P(1) = a{r^{1 - 1}} = \frac{{a({r^1} - 1)}}{{r - 1}} \Rightarrow a = a{/tex}

{tex}\therefore {/tex} P(1) is true

Let P(n) be true for n = k

{tex}\therefore P(k) = a + ar + a{r^2} + .... + a{r^{k - 1}}{/tex}{tex} = \frac{{a({r^k} - 1)}}{{r - 1}}{/tex} .... (i)

For n = k + 1

R.H.S. {tex} = \frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}

L.H.S. {tex} = \frac{{a({r^k} - 1)}}{{r - 1}} + a{r^k}{/tex} [ Using (i)]

{tex} = \frac{{a{r^k}}}{{r - 1}} - \frac{a}{{r - 1}} + a{r^k}{/tex}

{tex} = a{r^k} \cdot \left( {\frac{1}{{r - 1}}+1} \right) - \frac{a}{{r - 1}} = \frac{{a{r^{k + 1}}}}{{r - 1}} - \frac{a}{{r - 1}}{/tex}{tex} = \frac{{a{r^{k + 1}} - a}}{{r - 1}}{/tex}

{tex} = \frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}

{tex}\therefore {/tex} P(k + 1) is true

Thus P(k) is true {tex} \Rightarrow {/tex} P(k + 1) is true

Hence by principle of mathematical induction, P(n) is true for all {tex}n \in N{/tex}.

Posted by Tripti Verma 4 days, 8 hours ago

- 3 answers

Nitin Tiwari 3 days ago

Taking RHS side
Sin2x sin10x
Sin(6x-4x) sin(6x+4x)
(Sin6xcos4x - cos6xsin4x)(sin6xcos4x+cos6xsin4x) : (since-sin(A+B)=sinAcosB+sinBcosA)(sin(A+B)=sinAcosB- sinBcosA)
Sin^2(6x)cos^2(4x) - cos^2(6x)sin^2(4x)
Sin^2(6x)(1 - sin^2(4x)) - (1 - sin^2(6x))sin^2(4x)
Sin^2(6x) - sin^2(4x)sin^2(4x) - sin^2(4x) + sin^2(6x )sin^2(4x)
Sin^2(6x) - sin^2(4x)
Hence . Proved

Posted by Yamuna Shree Karpagarajan 5 days, 1 hour ago

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Posted by Ghanshyam Shukla 5 days, 5 hours ago

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Yash Goswami💞😎😎😎💞 5 days, 3 hours ago

Bro just solve rd sharma and u will get all types of questions of complex no.s

Posted by Prabhat Sahrawat 6 days, 1 hour ago

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