In \(\triangle ABC,\angle A+\angle B+\angle C=180^o\)(Angle sum property)
\(\Rightarrow 90^o+\angle B+\angle C=180^o(\angle A=90^o)\)
\(\Rightarrow \angle B+\angle C=90^o\)
\(\therefore \angle C=\angle ACB=90^o-\angle B...(i)\)
Also in \(\triangle ALB,\)
\(\angle ALB+\angle BAL+\angle B=180^o\)  (Angle sum property)
\(\Rightarrow 90^o+\angle BAL+\angle B=180^o\)
\(\angle BAL=90^o-\angle B....(ii)\)
Hence, from (i) and (ii), we get
\(\angle BAL=\angle ACB\)

\(\left( \sqrt { 3 } +1 \right) \left( 1-\sqrt { 12 } \right) +\frac { 9 }{ \left( \sqrt { 3 } +\sqrt { 12 } \right) } \)
\(=\left( \sqrt { 3 } -6+1-\sqrt { 12 } \right) +\frac { 9 }{ \left( \sqrt { 12 } +\sqrt { 3 } \right) } \times \frac { \left( \sqrt { 12 } -\sqrt { 3 } \right) }{ \left( \sqrt { 12 } -\sqrt { 3 } \right) } \)
\(=\left( \sqrt { 3 } -6+1-\sqrt { 12 } \right) +\frac { 9\left( \sqrt { 12 } -\sqrt { 3 } \right) }{ 12-3 } \)
\(=\sqrt { 3 } -5-\sqrt { 12 } +\sqrt { 12 } -\sqrt { 3 } \)
= -5

\(\frac { { \left( { x }^{ \frac { 1 }{ 2 } } \right) }^{ \frac { -2 }{ 3 } }{ \left( { y }^{ 4 } \right) }^{ \frac { 1 }{ 2 } } }{ { x }^{ \frac { -1 }{ 4 } }{ y }^{ \frac { -1 }{ 4 } } } ={ x }^{ \frac { -1 }{ 3 } }.{ y }^{ 2 }.{ x }^{ \frac { 1 }{ 4 } }.{ y }^{ \frac { 1 }{ 4 } }\)
\({ x }^{ \frac { -1 }{ 12 } }.{ y }^{ \frac { 9 }{ 4 } }=\frac { { y }^{ \frac { 9 }{ 4 } } }{ { x }^{ \frac { 1 }{ 12 } } } \)

3x=y+3; three solutions are x=1, y=0; x=2, y=3 and x=0, y=-3

From graph it is clear that line meets x-axis at(1,0) and y-axis at (0,-3)

Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To prove: (i) AD bisects BC (ii) AD bisects \(\angle A\)
Proof: (i) In right \(\triangle ADB\) and right \(\triangle ADC\)
Hyp.AB = Hyp. AC
Side AD = Side AD

\(\triangle ADB\cong \triangle ADC\) | RHS rule
BD = CD | C.P.C.T
AD bisects BC
(ii) \(\triangle ADB\cong \triangle ADC\)
\(\angle BAD=\angle CAD\)
AD bisects \(\angle A\)

Construction: Draw ray BL \(\bot \) PQ and ray CM \(\bot \) RS.

BL \(\bot \) PQ,CM \(\bot \) RS and PQ || RS BL||CM
\(\angle \) LBC=\(\angle \) MCB
|Alternate Interior Angles
\(\angle\)ABL=\(\angle\)LBC
Angle of incidence= Angle of reflection
\(\angle \)MCB=\(\angle \)MCD
Angle of incidence= Angle of reflection
From (1),(2) and (3) we get
\(\angle \)ABL =\(\angle \)MCD
Adding (1) and (4) , we get
\(\angle \) LBC+\(\angle \)ABL =\(\angle \)MCB+\(\angle \)MCD
\(\Rightarrow \) \(\angle \)ABC=\(\angle \)BCD
But these form a pair of equal alternate interior angles
So AB || CD.

-2x+3y=6
Comparing with ax+by+c=0, we get
a=2, b=3, c=-6

I

\(p\left( -\frac { 1 }{ 3 } \right) =3\left( -\frac { 1 }{ 3 } \right) +1=-1+1=0\)
\(\therefore -\frac { 1 }{ 3 } \) is a zero of p(x)

quadratic

Coefficient of x² = -1

\(\frac { 291 }{ 99 } \)

\(\frac { 3 }{ 13 } \)= 0.230769230769...=\(0.\overline { 230769 } \)

There can be infinitely many rational numbers between 3 and 4.
\(\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\\ \frac { 3+\frac { 7 }{ 2 } }{ 2 } =\frac { 13 }{ 4 } \)
\(\\ \frac { 3+\frac { 13 }{ 4 } }{ 2 } =\frac { 25 }{ 8 }\)
\( \\ \frac { 3+\frac { 25 }{ 8 } }{ 2 } =\frac { 49 }{ 16 } =\frac { 3+\frac { 49 }{ 16 } }{ 2 } =\frac { 97 }{ 32 } =\frac { 3+\frac { 97 }{ 32 } }{ 2 } =\frac { 193 }{ 64 } \)
Thus, six rational numbers between 3 and 4
\(\frac { 193 }{ 64 } ,\frac { 97 }{ 32 } ,\frac { 49 }{ 16 } ,\frac { 25 }{ 8 } ,\frac { 13 }{ 4 } \)and \(\frac { 7 }{ 2 } \)
Aliter
\(3=\frac { 3 }{ 1 } =\frac { 3\times 7 }{ 1\times 7 } =\frac { 21 }{ 7 } \)
\(\\ 4=\frac { 4 }{ 1 } =\frac { 4\times 7 }{ 1\times 7 } =\frac { 28 }{ 7 } \)
6 + 1 = 7
the six rational numbers between 3 and 4 can be taken as 
\(\frac { 22 }{ 7 } ,\frac { 23 }{ 7 } ,\frac { 24 }{ 7 } ,\frac { 25 }{ 7 } ,\frac { 26 }{ 7 } \) and \(\frac { 27 }{ 7 } \)