(i) \(\frac { -4 }{ 11 } \)

Thus we see that \(\frac { -4 }{ 11 } =-0.\overline { 36 } \)
(ii) \(\frac { 11 }{ 75 } \)

Thus we see that \(\frac { 11 }{ 75 } =0.14\overline { 6 } \)
A rational number can be expressed by 
(i) either a terminating
(ii) or a non-terminating and recurring (repeating) decimal expansion. 
The converse of this statement is also true. That is, if the decimal expansion of a number is terminating or non-terminating and recurring, then the number is a rational number.

Number of zeros = 1 (The curve is intersecting at the origin)

P(x) = 2x +5
= \(2\left( x+\frac { 5 }{ 2 } \right) \)
\(p\left( -\frac { 5 }{ 2 } \right) =2\left( -\frac { 5 }{ 2 } \right) +\frac { 5 }{ 2 } \)
= -5 + 5 = 2(0) = 0
Hence \(-\frac { 5 }{ 2 } \) is the zero of p(x)

(i) A = {a, e, i, o, u}
(ii) B = {b, c, e, o}
(iii) A – B = {a, i, u}
(iv) B – A = {b, c}
(v) A′ = {b, c, d, g}
(vi) B′ = {a, d, g, i, u}
(vii) U = {a, b, c, d, e, g, i, o, u}.

\(\frac { 1 }{ { x }^{ 2 } } +3x-4\) is not a polynomial.
Since the exponent of x² is not a whole number but it is \(\left( \frac { 1 }{ 2 } ={ x }^{ -2 } \right) \) negative number

U = {c, d, e, f, g, h, i, j}, A = {c, d, g, j}
A′ = {e, f, h, i}

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

Supplement of Right angle (90°) = 180° - 90°
= 90°

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

To multiply (4x – 5) and (2x² + 3x – 6) distribute each term of the first polynomial to every term of the second polynomial. In this case, we need to distribute the terms 4x and –5. Then gather the like terms and combine them:

= 8x³+12x²-24x+10x²-15x+30
= 8x³ +2x²–39x +30
Aliter: You may also use the method of detached coefficients:

∴  (4x –5) (2x²+3x–6) = 8x³ + 2x² – 39x + 30

(i) To plot (2, 4), draw a vertical line at x = 2 and draw a horizontal line at y = 4. The intersection of these two lines is the position of (2, 4) in the Cartesian plane. Thus, the Point A (2, 4) is located in the I quadrant of Cartesian plane.
(ii) To plot (–3, 5), draw a vertical line at x = –3 and draw a horizontal line at y = 5. The intersection of these two lines is the position of (–3, 5) in the Cartesian plane. Thus, the Point B (–3, 5) is located in the II quadrant of Cartesian plane.
(iii) To plot (–4, –5), draw a vertical line at x = –4 and draw a horizontal line at y =  -5. The intersection of these two lines is the position of (-4, -5) in the Cartesian plane. Thus, the Point C (–4, –5) is located in the III quadrant of Cartesian plane.
(iv) To plot (4, –2), draw a vertical line at x = 4 and draw a horizontal line at y = –2. The Intersection of these two lines is the position of (4,–2) in the Cartesian plane. Thus, the Point D (4,–2) is located in the IV quadrant of Cartesian plane

\(\therefore\) The Quotient : 4x²+2x+1
The Remainder : 0

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-y_{ 1 })^{ 2 } } \)
AB= \(\sqrt { (-2-6)^{ 2 }+(-4+4)^{ 2 } } \)
= \(\sqrt { (-8)^{ 2 }+0 } =\sqrt { 64 } =8\)
BC = \(\sqrt { (2+2)^{ 2 }+(10+4)^{ 2 } } \)
= \(\sqrt { (4)^{ 2 }+(14)^{ 2 } } =\sqrt { 16+196 } =\sqrt { 212 } \)
AC = \(\sqrt { (2-6)^{ 2 }+(10+4)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(14)^{ 2 } } =\sqrt { 1+-196 } =\sqrt { 212 } \)
BC = AC =\(\sqrt { 212 } \) (Two sides are equal)
∴  ABC is an isosceles triangle.

i) Draw a line and mark a point A on it.
ii) Mark a point B such that AB = 9.3 cm.
iii) Mark a point C on this line such that BC = 1 unit
iv) Find the midpoint of AC by drawing perpendicular bisector of AC and let it be O
v) With O as center and OC = OA as radius, draw a semicircle
vi) Draw a line BD, which is perpendicular to AB at B.
vii) Now BD = \(\sqrt { 9.3 } \)  which can be marked in the number line as the value of  BE = BD =\(\sqrt { 9.3 } \)

In the given figure
AB = AC (Given)
BM = MC (AM is the median of the \(\triangle\)ABC)
AM is common (By SSS congruency)
\(\therefore\triangle ABM\cong\triangle ACM\)