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let 3 and root 5 is irratonal no and 1 upon 3 is also irratonal no then multiply it to 3 root 5 then root 5 is required t and we know that root five is irrational mo

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If alpha and beta are zeroes of polynomial p (x) x^2-x-m such that alpha - beta = a . Theb find the value of m.

Posted by Vanshika Rajput 6 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

Since {tex}\alpha \text { and } \beta{/tex} are the zeroes of the polynomial, then
{tex}{/tex}{tex}\alpha + \beta = - \frac { \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } }{/tex}{tex}{/tex}
{tex}{/tex}{tex}\Rightarrow\ \alpha + \beta = - \left( \frac { - 1 } { 1 } \right) = 1{/tex}.........(i)
Given, {tex}\alpha - \beta = 9{/tex}...............(ii)
Solving (i) and (ii), {tex}\alpha = 5 , \beta = - 4{/tex}
{tex}\alpha \beta = \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }{/tex}
{tex}\Rightarrow\ \alpha \beta = - m{/tex}
{tex}\Rightarrow\ {/tex}(5)(-4) = -m
{tex}\Rightarrow{/tex}m = 20
So,required value of m is 20

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3x4-16x3+26x2-16x+3=0

Posted by Sanjeev Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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If tan = root 1-a^2 then prove that sec+tan^3cosec =( 2-a^2)^3/2

Posted by Yachna Sharma 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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If a polynomial x•4-3x•3-6x•2+kx-16 is exactly divisible by x•3-3x+2, find the value of K.

Posted by Sanju Dhami 6 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

Given p(x) = x⁴- 3x³- 6x² + kx - 16
and g(x) = x²- 3x + 2

On long division of f(x) by g(x) we get

Now, remainder = (k - 24)x
When p(x) is exactly divisible by g(x), then remainder = 0
{tex}\Rightarrow{/tex} (k - 24)x = 0
k - 24 = 0
k = 24

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For the highest common factor 52 and 117 and also find the value of x and y if it express in the form 52x + 117y.

Posted by Khushi Nagar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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Use euclid division algorithm to find the largest number which divides 957 and 1280 leaving 5 in each case.

Posted by Khushi Nagar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

Given numbers are 957 and 1280 and remainder is 5 in each case.Then , new numbers after subtracting remainders are
957 – 5 = 952 and 1280 – 5 = 1275
Now, by using Euclid's Division lemma , we get
1275 = (952 × 1) + 323
Here remainder = 323
So, on taking 952 as dividend and 323 as new divisor and then apply Euclid's Division lemma, we get
952 = (323 × 2) + 306
Again, remainder = 306.

So, on taking 323 as dividend and 306 as new divisor and then apply Euclid's Division lemma, we get
323 = (306 × 1) + 17
Again, remainder = 17.

So, on taking 306 as dividend and 17 as new divisor and then apply Euclid's Division lemma, we get
306 = (17 × 18) + 0
Here, remainder = 0.

Since, remainder has now become zero and the last divisor is 17.

Therefore, HCF of 952 and 1275 is 17.

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In an exam ,the no.of passed to failed is 3:1 .Had 8 more appeared and 6 less passed ,the ratio would become 2:1. How many students appeared in the exam?

Posted by Sudeepto Dixit 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sudeepto Dixit 7 years, 6 months ago

136

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Show that the square of any positive integer cannot be of form 5q+2 or 5q +3 for any integer q

Posted by Khushi Nagar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get
{tex}\style{font-family:Arial}{\begin{array}{l}n=5q+1,5q+2,5q+3\;and\;5q+4\;\\\end{array}}{/tex}
Now (5q)² = 25q² = 5m, where m = 5q², which is an integer;
{tex}\style{font-family:Arial}{\begin{array}{l}(5q\;+\;1)^{\;2}\;=\;25q^2\;+\;10q\;+\;1\;=\;5(5q^2\;+\;2q)\;+\;1\;=\;5m\;+\;1\\where\;m\;=\;5q^2\;+\;2q,\;which\;is\;an\;integer;\\\;(5q\;+\;2)^2\;=\;25q^2\;+\;20q\;+\;4\;=\;5(5q^2\;+\;4q)\;+\;4\;=\;5m\;+\;4,\\\;where\;m\;=\;5q^2\;+\;4q,\;which\;is\;an\;integer;\\\;(5q\;+\;3)^{\;2}\;=\;25q^2\;+\;30q\;+\;9\;=\;5(5q^2\;+\;6q+\;1)\;+\;4\;=\;5m\;+\;4,\\\;where\;m\;=\;5q^2\;+\;6q\;+\;1,\;which\;is\;an\;integer;\\\;(5q\;+\;4)^2\;=\;25q^2\;+\;40q\;+\;16\;=\;5(5q^2\;+\;8q\;+\;3)\;+\;1\;=\;5m\;+\;1,\;\\where\;m\;=\;5q^2\;+\;8q\;+\;3,\;which\;is\;an\;integer\\\end{array}}{/tex}
Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.
It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

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What is the value of cos square 67 - sin square 23

Posted by Astha Biswas 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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Astha Biswas 7 years, 5 months ago

Thnks

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Nikhil Gupta 7 years, 6 months ago

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Show that any number of the form (6)x, x€N can never end with the digit 0

Posted by Sanhji Agarwal 5 years, 8 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 6 months ago

{tex}6 ^ { n } = ( 2 \times 3 ) ^ { n } = 2 ^ { n } \times 3 ^ { n }{/tex}
{tex}\therefore \quad 5 \text { is not a factor of } 6 ^ { n }{/tex}
{tex}\therefore {/tex} It never ends with 0.

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Application of mathematics in other fields

Posted by Vishnu Soni 7 years, 6 months ago

CBSE > Class 10 > Mathematics