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Yogita Ingle 6 years, 11 months ago
x + a is factor x = -a
2x2 + 2ax + 5x + 10 =0
2(-a)2 + 2 a(-a) + 5(-a) + 10 = 0
2a2 - 2a2 - 5a + 10 = 0
- 5a + 10 = 0
- 5a = -10
a = 10/5
a = 2
Posted by Gaurav Bajaj 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We know that, Sin30° =(1/2), Cos45°=(1/√2), tan60°=√3, putting these values in the given expression, we get :-
{tex}2{\sin ^2}30^\circ - 3{\cos ^2}45^\circ + {\tan ^2}60^\circ {/tex}
{tex} = 2{\left( {\frac{1}{2}} \right)^2} - 3\left( {\frac{1}{{\sqrt 2 }}} \right)^2+ {\left( {\sqrt 3 } \right)^2}{/tex}
{tex} = \frac{2}{4} - \frac{3}{2} + 3{/tex}
{tex} = \frac{{2 - 6 + 12}}{4}{/tex}
{tex} = \frac{8}{4} = 2{/tex}
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Yogita Ingle 6 years, 11 months ago
First find length of each sides of ∆
Let A( 4 , 0) B(0, 0) and C (0 , 3)
use distance formula,
AB =√(4²+0) =4
BC= √(0+3²) = 3
CA =√(4²+3²) =5
Now , perimeter of ∆ = 3 + 4 + 5= 12 unit
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Sia ? 6 years, 6 months ago
If a and b are odd numbers then it should be in 2q+1 or 2q+3 form where q is a positive integer.
Let a = 2q + 3 , b = 2q + 1 and a > b
Now, {tex}\frac{a + b } {2} = \frac{ 2q + 3 + 2q + 1}{2}{/tex}
{tex}= \frac { 4 q + 4 } { 2 }{/tex}
= 2q + 2
{tex}\frac{a+b}{2}{/tex}=2(q+1) = an even number..........(1)
Now
{tex}\frac { a - b } { 2 } = \frac { ( 2 q + 3 ) - ( 2 q + 1 ) } { 2 }{/tex}
{tex}= \frac { 2 q + 3 - 2 q - 1 } { 2 }{/tex}
{tex}\frac{a-b}{2}= \frac { 2 } { 2 }{/tex} = 1 = an odd number..........(2)
Hence From (1) and (2) {tex}\frac{a+b}{2}{/tex} and {tex}\frac{a-b}{2}{/tex} are even and odd numbers respectively
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