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Ask QuestionPosted by Uallasgowda K R Uallasgowda K R 6 years, 5 months ago
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Posted by Hemant Kumar Meerotha 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
{tex}\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}{/tex}, x {tex}\neq{/tex} 5, 7
{tex}\Rightarrow \frac{(x-4)(x-7)+(x-6)(x-5)}{(x-5)(x-7)}=\frac{10}{3}{/tex}
{tex}=\frac{x^{2}-11 x+28+x^{2}-11 x+30}{x^{2}-12 x+35}=\frac{10}{3}{/tex}
{tex}\Rightarrow{/tex} 3[2x2 - 22x + 58] = 10[x2 - 12x + 35]
{tex}\Rightarrow{/tex} 6x2 - 66x + 174 = 10x2 - 120x + 350
{tex}\Rightarrow{/tex} 4x2 - 54x + 176 = 0
{tex}\Rightarrow{/tex} 2x2 - 27x + 88 = 0
{tex}\Rightarrow{/tex} 2x2 - 16x - 11x + 88 = 0
{tex}\Rightarrow{/tex} 2x(x - 8) -11(x - 8) = 0
{tex}\Rightarrow{/tex} (2x - 11) (x - 8) = 0
{tex}\Rightarrow{/tex} x = {tex}\frac{{11}}{2}{/tex}, 8
Posted by Hemant Kumar Meerotha 6 years, 5 months ago
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Posted by Preeti Karki 6 years, 5 months ago
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Posted by Sangeeta Pandey 6 years, 5 months ago
- 1 answers
Yogita Ingle 6 years, 5 months ago
Step 1: Since 12576 > 4052, apply the division lemma to 12576 and 4052, to get
12576 = 4052 × 3 + 420
Step 2: Since the remainder 420 ≠ 0, apply the division lemma to 4052 and 420, to get
4052 = 420 × 9 + 272
Step 3: Consider the new divisor 420 and the new remainder 272, and apply the division lemma to get
420 = 272 × 1 + 148
Consider the new divisor 272 and the new remainder 148, and apply the division lemma to get
272 = 148 × 1 + 124
Consider the new divisor 148 and the new remainder 124, and apply the division lemma to get
148 = 124 × 1 + 24
Consider the new divisor 124 and the new remainder 24, and apply the division lemma to get
124 = 24 × 5 + 4
Consider the new divisor 24 and the new remainder 4, and apply the division lemma to get
24 = 4 × 6 + 0
Hence, the HCF of 12576 and 4052 is 4.
Posted by Yash Bhardwaj 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
3x - y = 3
9x - 3y = 9
The given pair of linear equations is
3x - y = 3..............(1)
9x - 3y = 9.............(2)
From equation(1),
y = 3x - 3...................(3)
9x - 3(3x - 3) = 9
{tex}\Rightarrow{/tex} 9x - 9x + 9 = 9
{tex}\Rightarrow{/tex} 9 = 9
which is true. Therefore, equation (1) and (2) have infinitely many solutions.
Posted by Swastirajmoharana Newton 6 years, 5 months ago
- 1 answers
Nm ???? 6 years, 5 months ago
Posted by Gourav Dhull 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The given equations may be written as
{tex} \frac { x + 1 } { 2 } + \frac { y - 1 } { 3 } = 8 {/tex}
⇒ 3(x + 1) + 2 (y - 1) = 48
⇒ 3x + 3 + 2y - 2 = 48
⇒ 3x + 2y + 1= 48
⇒ 3x+2y = 47 ... (i)
{tex}\frac { x - 1 } { 3 } + \frac { y + 1 } { 2 } = 9{/tex}
⇒ 2(x -1 ) + 3(y +1) = 54
⇒ 2x - 2 + 3y + 3 = 54
⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53. ... (ii)
Multiplying (i) by 2 and (ii) by 3 and subtracting, we get
(4 - 9)y = 94-159
{tex} \Rightarrow - 5 y = - 65 {/tex}
{tex}\Rightarrow y = \frac { - 65 } { - 5 } {/tex}
{tex}\Rightarrow y = 13{/tex}
Putting y = 13 in (i), we get
3x + (2 {tex} \times{/tex} 13) = 47
{tex} \Rightarrow{/tex} 3x + 26 = 47
{tex} \Rightarrow{/tex} 3x = (47 - 26)
{tex} \Rightarrow{/tex}3x = 21
{tex} \Rightarrow \quad x = \frac { 21 } { 3 } = 7{/tex}
So, x = 7
Hence, x = 7 and y = 13
Posted by Omesh Sarna 6 years, 5 months ago
- 2 answers
Nm ???? 6 years, 5 months ago
Posted by Sanjib Dey 6 years, 5 months ago
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Posted by Durgesh Patel 6 years, 5 months ago
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Nm ???? 6 years, 5 months ago
Posted by Rajeev Singh 6 years, 5 months ago
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Gaurav Seth 6 years, 5 months ago
To prove that √5 is irrational number
Let us assume that √5 is rational
Then √5 =
(a and b are co primes, with only 1 common factor and b≠0)
⇒ √5 =
(cross multiply)
⇒ a = √5b
⇒ a² = 5b² -------> α
⇒ 5/a²
(by theorem if p divides q then p can also divide q²)
⇒ 5/a ----> 1
⇒ a = 5c
(squaring on both sides)
⇒ a² = 25c² ----> β
From equations α and β
⇒ 5b² = 25c²
⇒ b² = 5c²
⇒ 5/b²
(again by theorem)
⇒ 5/b-------> 2
we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.
This contradiction arises because we assumed that √5 is a rational number
∴ our assumption is wrong
∴ √5 is irrational number
Posted by Ankit Kumar 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
In mathematics, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, 3 × 5 is a factorization of the integer 15, and is a factorization of the polynomial x² – 4.
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Sia ? 6 years, 5 months ago
Given: (sec{tex}\theta{/tex} + tan{tex}\theta{/tex}) = m and (sec{tex}\theta{/tex} - tan{tex}\theta{/tex}) = n,
LHS = mn = (sec{tex}\theta{/tex} + tan{tex}\theta{/tex})(sec{tex}\theta{/tex} - tan{tex}\theta{/tex})
= sec2{tex}\theta{/tex} - tan2{tex}\theta{/tex} = 1 = RHS [{tex}\because{/tex} sec2{tex}\theta{/tex} - tan2{tex}\theta{/tex} = 1]
{tex}\therefore \sqrt {mn} = 1{/tex}
1Thank You