We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects LC
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle)     .........(1)
.Also BC II AD and AC is a transversal.
∴  ㄥ2 = ㄥ4 (Alternate interior angle)  ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ BC = AB (side opposite to equal angles are equal) (4)
Similarly AD = DC            ....... (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram)        .......(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Let A(1, 1),B(5, 4) and C(-2, 5)
Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(5-1)^2+(4-1)^2}\)
\(=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\)
BC =\(\sqrt{(-2-5)^2+(5-4)^2}\)
\(=\sqrt{(-7)^2+1^2}=\sqrt{49+1}=\sqrt{50}\)
AC =\(\sqrt{(-2-1)^2+(5-1)^2}\)
\(=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

AB = 5, AC = 5,
\(\therefore\) ABC is an isosceles triangle .................(1)
BC² = AB² + AC²
50 = 25 + 25\(\Rightarrow\) 50 = 50
\(\therefore\) \(\angle\)A = 90° ...................(2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

In the given \(\triangle ABC\)
\(\angle A=64^0\ and\ \angle B=58^0\)
\(\angle C=180^0-(64^0+58^0)\)
= 180° - 122°
= 58°
Since OC is the bisector of \(\angle C\)
\(y=\frac{58^0}{2}=29^0\)
Given \(\triangle OBC\)
\(\angle OBC=\frac{58^0}{2}=29^0\)
\(\angle OCB=29^0\)
\(\therefore\angle BOC=180^0-(29^0+29^0)\)
x = 180° - 58°
x = 122°
\(\angle x=122^0\ and\ \angle y=29^0\).

The two rational numbers are 2.2362 and 2.2363 (It has many answers)

Let A be the required number to be subtracted
(y⁴ + 2y³ - 3y² + 8) - A = y⁴ - 2y³ + 6
y⁴ + 2y³ - 3y² + 8 - (y⁴ - 2y³ + 6) = A
y⁴ + 2y³ - 3y² + 8 - y⁴ + 2y³ - 6 = A
4y³-3y²+2 = A
Hence 4y³-3y²+2 must be subtracted.

(i) The algebraic expression is 3n + 1
(ii) The Algebraic expression is 3n - 2

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60°    (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

(i) A = {1,2,5,6,7}
(ii) B = {3,4,5,6}
(iii) A ∩ B = {5,6}
(iv) A U B = {1,2,3,4,5,6,7}
(v) A-B = {1,2,7}
(vi) B - A = {3,4}
(vii) (A-B) ∩ (B -A) = { }.

(3, 5)

Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A \(\cap\) B) = 50
n(A \(\cup\) B) = 500 .........(1)
n(A) + n(B) - n(A \(\cap\) B) = 400 + 200 - 50
= 600 - 50
= 550  ...........(2)
From (1) and (2) we get
\(n(A\cup B)\ne n(A) +n(B)-n(A\cap B)\)
\(\therefore\) The given data is incorrect.

Let x = 0.00010001 ............. \(\rightarrow\) (1)
10000x = 1.00010001 .......... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 10000x - x = 1.00010001 ......... (-)
                                   0.00010001 .........
                                 ____________
                     9999x = 1
      \(x=\frac{1}{9999}\)      

\( -5.132 =\frac{-5132}{1000} \\ =\frac{-1283}{250} \)

Distance AB = \(\sqrt { (a-a)^{ 2 }+(3+2)^{ 2 } } \)
= \(\sqrt { 0+{ 5 }^{ 2 } } =\sqrt { 25 } =5\)
BC = \(\sqrt { (a-a)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { 0+9 } =\sqrt { 9 } =3\)
AC = \(\sqrt { (a-a)^{ 2 }+(0+2)^{ 2 } } \)
= \(\sqrt { 0+2^{ 2 } } =\sqrt { 4 } =2\)
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear

Through T draw TE|| AB.
\(\therefore \angle BAT+\angle ATE=180^0(AB||TE)\)
\(140^0+\angle ATE=180^0\)
\(\angle ATE=180^0-140^0=40^0\)
Similarly \(\angle ETC+\angle TCD=180^0(TE||CD)\)
\(\angle ETC+150^0=180^0\)
\(\angle ETC=180^0-150^0=30^0\)
\(x=\angle ATE+\angle ETC\)
= 40° + 30° = 70°
x = 70°

\({7\over 16}={7\over 2^4}={7\over 2^4\times 5^6}\)
\(\therefore {7\over 16}\) has a terminating decimal expansion.

ㄥP = ㄥR= 168⁰

(i) {5,9}
(ii) {1,2,3,4,5,7,8,9}

The set B is a Null set

{C, Z, E, H, O, S, L, V, A, K, I}

10x + 9 = 0
10x = - 9
x = \(-\frac { 9 }{ 10 } \)
\(-\frac { 9 }{ 10 } \) is the root of the polynomial

It is not a trapezium

It is a parallelogram

The line is on the y - axis