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Ask QuestionPosted by Shahareyar Anjum Kha Sufiyan 7 years, 6 months ago
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Posted by Abhishek Patel 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let p(x) = ax3 + 4x2 + 3x - 4 and q(x) = x3 - 4x + a be the given polynomials. The remainders when p(x) and q(x) are divided by (x - 3) are p(3) and q(3) respectively.
By the given condition, we have
p(3) = q(3)
{tex}\Rightarrow{/tex} a {tex}\times{/tex}33 + 4 {tex}\times{/tex}32 + 3 {tex}\times{/tex}3 - 4 = 33 - 4 {tex}\times{/tex}3 + a
{tex}\Rightarrow{/tex} 27a + 36 + 9 - 4 = 27 - 12 + a
{tex}\Rightarrow{/tex} 26a + 26 = 0 {tex}\Rightarrow{/tex} 26a = -26 {tex}\Rightarrow{/tex} a = -1
Posted by Navneet Singal 7 years, 6 months ago
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Posted by Jyoti Raghuvanshi 7 years, 6 months ago
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Charu Sharma 7 years, 6 months ago
Posted by Chandhu Y 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let N = 5q + 1
{tex}\therefore{/tex} N2 = (5q + 1)2
{tex}\Rightarrow{/tex} N2 = 25q2 + 10q + 1
N2= 5(5q2 + 2q) + 1
= 5m + 1, where q is soqe integer
Hence, the square of any positive integer of the form 5m + 1 will leave a remainder 1 when divided by 5 for some integer m .
Posted by Nalinikanta Sahoo 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
117 = 13 {tex}\times{/tex} 3 {tex}\times{/tex} 3
65 = 13 {tex}\times{/tex} 5
HCF (117, 65) = 13
LCM(117,65) = 13 {tex}\times{/tex} 5 {tex}\times{/tex} 3 {tex}\times{/tex} 3 = 585
Here is given that:
{tex}HCF =65m-117{/tex}
{tex}13=65m-117{/tex}
{tex}65m=130{/tex}
m = {tex}\frac { 130 } { 6 5 } ={/tex}2
Posted by Keshav Jagnani 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have, x = 4 and x = -3.
Then,
x - 4 = 0 and x + 3 = 0
{tex}\Rightarrow{/tex} (x - 4)(x + 3) = 0
{tex}\Rightarrow{/tex} x2 + 3x - 4x - 12 = 0
{tex}\Rightarrow{/tex} x2 - x - 12 = 0
This is the required quadratic equation
Posted by Samiran Karki Chettri 7 years, 6 months ago
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Nipun Goyal 7 years, 6 months ago
Posted by Mumaiz Peer 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
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Posted by Mercy Catrin 7 years, 6 months ago
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Posted by Ashtrix Official 7 years, 6 months ago
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Posted by Ashtrix Official 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
sin(135)
= sin (90 + 45) // sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
= sin(90)cos(45) + cos(90)sin(45)
= ( 1 x 1/√2) + (0 x 1/√2)
= 1/√2
= (√2)/2
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Shahareyar Anjum Kha Sufiyan 7 years, 6 months ago
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