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From where i can get the solution of activity work ?

Posted by Shubham Anand 6 years, 3 months ago

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https://brainly.com/question/add?entry=1642&task_content=f+%3A(3%2C7)%2C(4%2C9)%2C5%2C11)and+g%3A(3%2C7)%2C(4%2C11)%2C(5%2C11)+are+two+relations+from+a+to+b+in+the+arrow++diagram+above+1)+are+the+relations+f+and+g+are+functions+2)+writ+f+in+roster+form+3)+write+(x+belongs+to+A+f+(x)%3D+g(x)+)+in+roster+form

Posted by Seema Sreekumar 6 years, 3 months ago

CBSE > Class 11 > Mathematics

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In a group of students half the no. Of students know hindi 2/3 of them know english 10 know both languages and 6 student do not know either english or hindi fi nd how many students are there in the groul

Posted by Shagun Chadak 6 years, 3 months ago

CBSE > Class 11 > Mathematics

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Sin3A+sin2A-sinA=4sinAcosA/2cos3A/2

Posted by Disha Goutam 6 years, 3 months ago

CBSE > Class 11 > Mathematics

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Construct five different examples of sets. Describe each with the help of a proposition

Posted by Subhalaxmi Sandha 6 years, 3 months ago

CBSE > Class 11 > Mathematics

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If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2+-√3. find other zeroes.

Posted by Hrithik Singh 6 years, 3 months ago

CBSE > Class 11 > Mathematics

2 answers

Sia ? 6 years, 3 months ago

Two zeros are {tex}2\pm\sqrt3{/tex}
Sum of Zeroes {tex}2 + \sqrt { 3 } + 2 - \sqrt { 3 } = 4{/tex}
and product of zeroes = {tex}( 2 + \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) = 4 - 3 = 1{/tex}
Hence quadratic polynomial formed out of this will be a factor of given polynomial,
So, x² - (sum of zeroes)x + product of zeroes
= x² - 4x + 1 will be a factor of given polynomial,
Divide given polynomial with x² - 4x + 1 to get other zeroes.

Now,
x² -2x - 35
= x² - 7x + 5x - 35
= x(x - 7) + 5(x - 7)
= (x - 5) (x - 7)
{tex}\therefore {/tex} Zeros are
x = 7 and x = -5
{tex}\therefore {/tex} Other two zeros are 7 and -5

2Thank You

Shantanu Kapse 6 years, 3 months ago

Let the remaining two zeros be X and Y, As, Sum of roots =6 , product of roots = -35 We get 2 equations, 2+√3 + 2-√3 + x +y = 6 x+y= 2-----(1) (2+√3)(2-√3)(xy) = -35 xy=-35 x=-35/y -----(2) Subs (2) in (1) y-35/y =2 y²-2y-35=0 y=[2±√(144)]/2 y=7 y=-5 x=-5 x=7 Hence the roots are (2±√3),(-5),(7) .

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Evaluate cos105degree+sin105degree

Posted by Muskan Chhilar 6 years, 3 months ago

CBSE > Class 11 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

The value of cos 105^o + sin 105^o is {tex}\frac{1}{\sqrt{2}}{/tex}
Cos trigonometric values are positive in “first and fourth” quadrants in coordinate axes.
Sin trigonometric values are positive in “first and second” quadrants in coordinate axes.
cos 105^o + sin 105^o = cos (90^o + 15^o) + sin (90^o + 15^o)
= - sin (15^o) + cos (15^o) (ascos (90^o + {tex}\theta{/tex}) = - sin {tex}\theta{/tex} and sin (90^o + {tex}\theta{/tex}) = cos {tex}\theta{/tex})
= cos 15^o- sin 15^o
{tex}\left.=\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos 15^{\circ}-\frac{1}{\sqrt{2}} \sin 15^{\circ}\right] \quad \text { (Multiply and divide by } \sqrt{2}\right){/tex}
{tex}=\sqrt{2}\left[\sin 45^{\circ} \cos 15^{\circ}-\cos 45^{\circ} \sin 15^{\circ}\right] \quad\left(A s \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right){/tex}
{tex}=\sqrt{2}\left[\sin \left(45^{\circ}-15^{\circ}\right)\right] \quad(A \sin (A-B)=\sin A \cos B-\cos A \sin B){/tex}
{tex}=\sqrt{2}\left[\sin 30^{\circ}\right]{/tex}
{tex}=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}{/tex}