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Ask QuestionPosted by Sudha Sudha 6 years, 9 months ago
- 2 answers
Posted by Deepanshu Kumar 6 years, 9 months ago
- 0 answers
Posted by Ujjwal Pradhan 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
HCF of 468 and 222
468 = (222 x 2) + 24
222 = (24 x 9) + 6
24 = (6 x 4) + 0
Therefore, HCF = 6
6 = 222 - (24 x 9)
= 222 - {(468 – 222 x 2) x 9 [where 468 = 222 x 2 + 24]
= 222 - {468 x 9 – 222 x 2 x 9}
= 222 - (468 x 9) + (222 x 18)
= 222 + (222 x 18) - (468 x 9)
= 222[1 + 18] – 468 x 9
= 222 x 19 – 468 x 9
= 468 x -9 + 222 x 19
Hence, HCF of 468 and 222 in the form of 468x + 222y is 468 x -9 + 222 x 19.
Posted by Tirth Patel 6 years, 9 months ago
- 3 answers
Posted by Riya Rai 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
a = bq + r , where 0≤ r <b
Explanation: Thus, for any pair of two positive integers a and b; the relation a = bq + r , where 0 ≤ r < b
will be true where q is some integer.
Posted by Zeeshan Khan 6 years, 9 months ago
- 1 answers
Gaurav Seth 6 years, 9 months ago
a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).
Posted by Monish Narayanan 6 years, 7 months ago
- 1 answers
Sia ? 6 years, 7 months ago
g(x)=a(x2+1)-x(a2+1)
= ax2 + a - a2x - x
=ax2 - a2x- x+a
= ax(x - a) - (x - a)
=(x-a)(ax-1)
g(x)=0 if x-a=0 or ax-1=0
Hence zeros of the polynomials are: a and {tex} \frac{1}{a}{/tex}
now,
Sum of the zeros ={tex}\mathrm a+\frac1{\mathrm a}=\frac{\mathrm a^2+1}{\mathrm a}=-\frac{-(\mathrm a^2+1)}{\mathrm a}=-\frac{\mathrm B}{\mathrm A}{/tex}
Product of zeros {tex} = \frac{1}{a}\times a{/tex}=1={tex}\frac aa=\frac CA{/tex}
Hence, the relationship is verified.
Posted by Shania Goria 6 years, 9 months ago
- 4 answers
Kunal Nawariya 6 years, 9 months ago
Kunal Nawariya 6 years, 9 months ago
Kunal Nawariya 6 years, 9 months ago
Posted by Aditya Yadav 6 years, 7 months ago
- 1 answers
Sia ? 6 years, 7 months ago
The zeros of f(x) are given by f(x) = 0
{tex} \Rightarrow {/tex}x2 + 7x + 12 = 0
{tex}\Rightarrow {/tex}(x + 4)(x + 3) = 0
{tex}\Rightarrow {/tex} x + 4 = 0 or, x + 3 = 0
{tex}\Rightarrow {/tex} x = -4 or, -3
Thus, the zeroes of f(x) = x2 +7x + 12 are {tex}\alpha = - 4 \text { and } \beta = - 3{/tex}
Posted by Manoj Jha 6 years, 9 months ago
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Posted by Manisha Choudhary 6 years, 9 months ago
- 2 answers
Posted by Durgesh Kumar Meghwanshi 6 years, 9 months ago
- 1 answers
Posted by Ajay Rathor 6 years, 9 months ago
- 1 answers
Posted by ????? ?????? 6 years, 9 months ago
- 3 answers
????? ?????? 6 years, 9 months ago
Isheta Singh 6 years, 9 months ago
Posted by Tinku Munda 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
So, HCF(441, 63) = 63
So, HCF (693, 567) = 63
441 = 63 x 7 + 0
Hence, HCF (693, 567, 441) = 63</div>
Posted by Anita Bhowmick 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
A quadratic polynomial have at most two zeros because the degree of x is 2.
Posted by P Yadav 5 years, 9 months ago
- 0 answers
Posted by P Yadav 6 years, 9 months ago
- 1 answers
Ziddi Akshit?????? ??? 6 years, 9 months ago
Posted by Aman Singh Rana G 6 years, 9 months ago
- 2 answers
Yogita Ingle 6 years, 9 months ago
The real numbers which are not rational i.e. which cannot be expressed in the form of <nobr>p/q</nobr> , where p and q are integers and q ≠ 0 are known as irrational numbers. The decimal expansion of an irrational number is neither terminating nor recurring.
For Example: √2, √3, <nobr>π</nobr> , etc.
Shalbi Kumari 6 years, 9 months ago
Posted by Vaibhav Sharma 6 years, 9 months ago
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Posted by Ramanand Soni 6 years, 9 months ago
- 0 answers
Posted by Teja Sri 6 years, 9 months ago
- 0 answers
Posted by Gaurav Kumar 6 years, 9 months ago
- 3 answers
Posted by Ŕòmàñçè Ķìñģ Sshìvàm 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 {tex}\times{/tex} 1+ 348
609 = 348 {tex}\times{/tex} 1 + 261
348 = 216 {tex}\times{/tex} 1 + 87
261 = 87 {tex}\times{/tex} 3 + 0.
Therefore, H.C.F. of 957 and 609 is = 87.
Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87.
Posted by Sahini Sharma 6 years, 9 months ago
- 0 answers
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