(i) \(3\over 4\) = 0.75

(ii) \(5\over 8\) = 0.625

(iii) \(9\over 25\) = 0.36

(X - Y) U (Y - X)

A' = {a, c, e, g}
(A')' = U - A'
= {a, b, c, d, e, f, g, h} - {a, c, e, g}
= {b, d, f, h}

p(x) = 2x³ - l1x² - 4x + 3
g(x) = 2x + 3
p(x) = 2x³ - 11x² - 4x + 3
\(p\left( \frac { -3 }{ 2 } \right) =2\left( \frac { -3 }{ 2 } \right) ^{ 3 }-11\left( \frac { -3 }{ 2 } \right) ^{ 2 }-4\left( \frac { -3 }{ 2 } \right) +3\)
= \(2\left( \frac { -27 }{ 8 } \right) -11\left( \frac { 9 }{ 4 } \right) +\frac { 12 }{ 2 } +3\)
= \(-\frac { 54 }{ 8 } -\frac { 99 }{ 4 } +6+3\)
= \(-\frac { 27 }{ 4 } -\frac { 99 }{ 4 } +9\)
= \(\frac { 27-99+36 }{ 4 } \)
= \(\frac { 27-99+36 }{ 4 } =\frac { -126+36 }{ 4 } \)
= \(-\frac { 90 }{ 4 } =-\frac { 45 }{ 2 } \)
= \(p\left( \frac { -3 }{ 2 } \right) \)\(\neq \) 0
P(x) is not a multiple of g(x)

Distance between the points (1, 2) and (4, 3)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (4-1)^{ 2 }+(3-2)^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } \)
= \(\sqrt { 9+1 } =\sqrt { 10 } \) = units

5x – 3 = 0 (or) 5x = 3
Then, x = \(\frac{3}{5}\)

A = {b, d, e, g, h}, B = {a, e, c, h}
A – B = {b, d, g}
n(A – B) = 3 .......... (1)
A ∩ B = {e, h}
n(A ∩ B) = 2 , n(A) = 5
n(A) – n(A∩B) = 5 - 2
= 3 ....... (2)
Form (1) and (2) we get n(A – B) = n(A) – n(A∩B).

\(\angle AOD+\angle DOC+\angle COB=180^0\) (angle of a straight line)
3x° + 90° + 2x° = 180°
5x° + 90° = 180°
5x° = 180° - 90°
\(x=\frac{90}{5}=18^0\)
The value of x = 18°.

ASSESSMENT
A= {A, S, E, M, N, T}

The x- coordinate is positive and y – coordinate is negative. So, Point(3,–8) lies in the IV quadrant.

Let p(x) = x³- 9x² + 26x + k
By factor theorem,(x-3) is a factor pf p(x),if p(3) = 0
p(3) = 0
3³ -9(3)² + 26(3) + k = 0
27 - 81 + 78 + k = 0
k = -24

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.

0.86 = \({8\over 10}+{6\over 100}={80+6\over 100}\)
        = \(86\over 100\) (or) \(43\over 50\)

We see that p(x) – q(x) is also a polynomial. Hence the subtraction of any two polynomials is also a polynomial.

The distance between the points (-4, 3), (2, -3) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2+4 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( { 6 }^{ 2 }+\left( -6 \right) ^{ 2 } \right) } =\sqrt { \left( 36+36 \right) } \)
\(=\sqrt { 36\times 2 } \)
\(=6\sqrt { 2 } \)

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, Band C.


Circum radius = 4.3 cm.

Let the people who like tea be T. 
Let the people who like coffee be C 
By using formula 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10 
The number of people who like both coffee and tea = 10.
(ii) The number of people who do not like tea 
n(T) = n(U) – n(T) = 45 – 35 = 10
(iii) The number of people who do not like coffee 
n(C) = n(U) – n(C) = 45 – 20 = 25.

The two given points are (5,-2), (1, a) and d = 5.
By distance formula
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 1-5 \right) ^{ 2 }+\left( a+2 \right) ^{ 2 } } =5\)
\(=\sqrt { 16+\left( a+2 \right) ^{ 2 } } =5\)
16 + (a+2)² = 25 (By squaring on both the sides)
(a+2)² = 25–16
(a+2)² = 9
(a+2) = \(\pm \) 3 (By taking the square root on both side)
a = –2 \(\pm \) 3
a = –2 + 3 (or) a = –2 –3
​​​​​​​a = 1 or –5.