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Ask QuestionPosted by Siddhant Raj 7 years, 8 months ago
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Posted by Inder Preet 7 years, 8 months ago
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Posted by Ayushi Yadav 7 years, 8 months ago
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Posted by Anand Raj 7 years, 8 months ago
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Posted by Vikram Kumar 7 years, 8 months ago
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Shagufta Nargis 7 years, 8 months ago
Posted by Lakshay Gautam 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The given pair of the equation is
{tex}\frac{4}{x} + 3y = 14{/tex} ....(1)
{tex}\frac{3}{x} - 4y = 23{/tex} ....(2)
Put {tex}\frac{1}{x} = x{/tex} ....(3)
Then equation (1) and (2) can be rewritten as
4x + 3y = 14 ...(4)
3x - 4y = 23 ...(5)
From equation (5),
4y = 3x - 23
{tex} \Rightarrow \;y = \frac{{3x - 23}}{4}{/tex} ....(6)
Substituting this value of y in equation (4) we get
{tex}4x + 3\left( {\frac{{3x - 23}}{4}} \right) = 14{/tex}
{tex} \Rightarrow{/tex} 25x = 56 + 69 = 125
{tex}\Rightarrow \;x = \frac{{125}}{{25}} = 5{/tex} ....(7)
Substituting this value of x in equation (6), we get
{tex}y = \frac{{3(5) - 23}}{4} = \frac{{15 - 23}}{4}{/tex}
{tex}= \frac{{ - 8}}{4} = - 2{/tex} ...(8)
From equation (3) and equation (7), we get {tex}\frac{1}{x} = 5{/tex}
{tex}\Rightarrow \;\frac{1}{x} = 5{/tex}
Hence, the solution of the given pair of the equation is,{tex}\frac{1}{x} = 5{/tex} y = -2.
Verification. Substituting {tex}\frac{1}{x} = 5{/tex}, y = -2,
We find that both the equations (1) and (2) are satisfied as shown below:
{tex}\frac{4}{x} + 3y = \frac{4}{{\left( {\frac{1}{5}} \right)}} + 3( - 2) = 20 - 6 = 14{/tex}
{tex}\frac{3}{x} - 3y = \frac{3}{{\left( {\frac{1}{5}} \right)}} - 4( - 2) = 15 + 8 = 23{/tex}
Hence, the solution is correct.
Posted by Khu Shi 7 years, 8 months ago
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Hemanshu Kumar 7 years, 8 months ago
Posted by Bharti Bhatia 7 years, 8 months ago
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Posted by Akash Meher 7 years, 8 months ago
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Posted by Vanshita Waghmare 7 years, 8 months ago
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Posted by Aniket Srivastava 7 years, 8 months ago
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Shourya Adhikari 7 years, 8 months ago
Posted by Robert Ronnie 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the supplementary angles be x° and y° (x° > y°)
Now, x° + y° = 180 ........ (i)
and x° - y° = 18 ..(ii)
From eqn. (ii), y° = x° - 18 ...(iii)
Put (iii) in (i)
x° + x° - 18° = 180°
⇒ 2x° = 180° + 18°
⇒ 2x° = 198°
⇒ x° = 99°
On substituting {tex}x ^ { \circ } = 99 ^ { \circ }{/tex} in eqn. (iii),
{tex}y ^ { \circ } = 99 ^ { \circ } - 18 ^ { \circ } = 81 ^ { \circ }{/tex}
{tex}\therefore {/tex}{tex}y ^ { \circ } = 81 ^ { \circ }{/tex}
Hence, the angles are {tex}99 ^ { \circ } \text { and } 81 ^ { \circ }{/tex}
Posted by Shubhi Sharma 7 years, 8 months ago
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Posted by Sudhansusekhar Mohanty 7 years, 8 months ago
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Posted by Rohit Saw Rohit 7 years, 8 months ago
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Posted by Anand Kumar 7 years, 8 months ago
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Posted by Suraj Bhandari 7 years, 8 months ago
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Posted by Jasmeet Kaur 7 years, 8 months ago
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Posted by Komal Soni 7 years, 8 months ago
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Akshay Kumar 7 years, 8 months ago
Posted by Niranchana Shree 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Total number of outcomes = 98
- Favourable outcomes are all numbers divisible by 8 i.e., 8, 16,24,....,96 =12 numbers
{tex}\therefore{/tex} Prob (integer is divisible by 8) = {tex}\frac{12}{98}{/tex} or {tex}\frac{6}{49}{/tex} - Prob(integer is not divisible by 8) = 1 - P(integer is divisible by 8) = 1-{tex}\frac{6}{46}{/tex} = {tex}\frac{43}{49}{/tex}
Posted by Agni Shikha 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.
1445 = 1190 × 1 + 255
1190 = 255 × 4 + 170
255 = 170 × 1 + 85
170 = 85 × 2 + 0
So, now the remainder is 0, then HCF is 85
Now,
85 = 255 - 170
= (1445 - 1190) - (1190 - 255 × 4)
= 1445 - 1190 - 1190 + 255 × 4
= 1445 - 1190 × 2 + (1445 - 1190) × 4
= 1445 - 1190 × 2 + 1445 × 4 - 1190 × 4
= 1445 × 5 - 1190 × 6
= 1190 × (- 6) + 1445 × 5
= 1190m + 1445n , where m = - 6 and n = 5
Posted by Anushka Yadav 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
n3 - n = n (n2 - 1) = n (n - 1) (n + 1)
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴ n (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)
Posted by Amarjeet Sharma 7 years, 8 months ago
- 1 answers
Posted by Amarjeet Sharma 7 years, 8 months ago
- 1 answers
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