3z-42^o=2z+13^o (Alternate interior angles)
z=42^o+13^o
z=55^o
\(\angle DNM=2z+13^o=110^o+13^o\)
=123^o
\(\angle CNM=180^o-\angle DNM\)(Linear pair)
=180^o-123^o=57^o

\(\frac { 1 }{ 2 } \)

(By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.
So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.
So, 6x² + 17x + 5 = 6x² + (2 + 15)x + 5
= 6x² + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)

63³

Required sum of probabilities=1

Mean \(=\frac { { m }x_{ 1 }+{ mx }_{ 2 }+...+{ mx }_{ n } }{ n } \)
 \(=m\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =m\overset { - }{ x } .\)

Class mark = \(\frac {90+120}{2}=105\)

Area = \(\frac { AB\times BC }{ 2 } =\frac { 4\times 4 }{ 2 } \)= 8 cm²

Formula

\(\angle PQR=90°\) 
\(\therefore\angle RPQ+\angle PQR=90°\)
But \(\angle RPQ=\angle PQR\)

\(\angle PQR=45°\)

A median of a triangle divides it into two triangles of equal areas.

Definition of an acute angle

2(2)+2=6

x-1=1-x ⇒ x=1 so a linear equation

\((1+3y)^3+(9y^2-1)\)
\(=(1+3y)^2+(3y+1)(3y-1)\)
\(=(1+3y)(1+3y+3y-1)\)
\(=6(1+3y)y\)

A polynomial of degree n has maximum number of terms as (n+1)

Given: BQ||CR
\(\because\) \(\Delta \) BCQ and \(\Delta \)BQR are on the same base BQ and between the same parallel BQ and CR.
\(\therefore\) ar(\(\Delta \)BCQ) = ar (\(\Delta \)BQR)    ......(i)
Also,  AP||BQ.     (Given)

Again, \(\Delta \)ABQ and \(\Delta \)PBQ are on the same base BQ and between the same parallels BQ and AP.
\(\therefore\) ar(\(\Delta \)ABQ) = ar (\(\Delta \)PBQ)    .......(ii)
Adding (1) and (2), we get
ar(\(\Delta \)BCQ) + ar(\(\Delta \)ABQ) = ar(\(\Delta \)BQR) + ar(\(\Delta \)PBQ)
ar(\(\Delta \)AQC) = ar(\(\Delta \)PBR).

Since LCM of 7 and 11 is 77,
ஃ \(\frac{5}{7}=\frac{5}{7}=\frac{11}{11}=\frac{55}{77}\)
and \(\frac{5}{7}=\frac{5}{7}=\frac{11}{11}=\frac{55}{77}\)
Hence, three rational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) are:
\(\frac{56}{77},\frac{57}{77},\frac{58}{77}\)

Modified Table 
[Minumum class-]

(ii) The class interval in which the maximum number of surname lie is 6-8.

Let the radius of the base of the cylinder be r cm.
= 5 cm
Lateral surface =94.2 cm²
\(\Rightarrow 2\pi rh=94.2\)
\(\Rightarrow 2\times 3.14 \times r\times 5=94.2\)
\(\Rightarrow r=\frac{94.2}{2\times3.14\times5}\)
\(\Rightarrow r=\frac{94.2}{31.4}\)
\(\Rightarrow r=3cm\)
(ii) r = 3cm,  h = 5 cm
ஃ Volume of the cylinder =\(\pi r^2h\)
= 3.14 x (3)² x 5 =141.3 cm³

Rs. 6705

Given: ABCD is a trapezium whose nonparallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE 11 AD.
Proof: ∵ AB || DE        I Given
and     AD || BE           I By construction
∴  Quadrilateral ABCD is a parallelogram.

∴ \(\angle BAD=\angle BED\)                        ....(1)
| Opp.\(\angle \) s of a || gm are equal
and    AD = BE                                 ...(2)
Opp. sides of a || gm are equal
But     AD = BC            ...(3) I Given
From (2) and (3),
BE = BC
∴ \(\angle BEC=\angle BCE\)        ....(4)
| Angles opposite to equal sides of a triangle are equal
\(\angle BEC+\angle BED=180°\)       | Linear Pair Axiom
⇒ \(\angle BCE+\angle BAD=180°\) I From (4) and (1)
⇒ Trapezium ABCD is cyclic.
I ∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic

2x+y=7
⇒ y=7-2x
Put x=0, we get y=7-2(0)=7-0=7
Put x=1, we get y=7-2(1)=7-2=5
Put x=2, we get y=7-2(2)=7-4=3
Put x=0, we get y=7-2(3)=7-6=1
Four solutions are (0,7), (1,5), (2,3) and (3, 1)

Trapezium

(i) \(B\rightarrow (-5,2)\)
(ii) \(C\rightarrow \left( 5,-5 \right) \)
(iii) E
(iv) G
(v) 6
(vi) -3
(vii) \(L\rightarrow (0,5)\)
(viii) \(M\rightarrow (-3,0)\)

This expression is not a polynomial in one variable because in the expression, three variables (x, y and t) occur.

Total number blood donors = 85
(i) Number of donors whose blood is useful for Cheeku = 3
\(\therefore \) Number of favourable events = 85 - 3 = 82
Now, required probability\(=\frac{82}{85}\)
(ii) Responsible, rationality, love
(iii) Blood donation helps the needy. In fact it is a great act of charity and co-operation.

(a) Probability that a student selected is opts French language = \(\frac{12}{50}=\frac{6}{25}\)
(b) Probability that a student selected does not opt for Japanese = 1-selected student opts Japnese
(c) Probability that selected either opts for Sanskrit or for German = Prob. of student opts Sanskrti+ Prob. of students opts German
\(\therefore \)No of student who opted German 
= 5 - (14 + 08 + 12 + 6)
= 50-40
= 10
\(\therefore \)Prob. that selected student either opt for Sanskrit or for German
\(=\frac{14}{50}+\frac{10}{50}\)
\(=\frac{24}{50}=\frac{12}{25}\)

In \(\Delta\)ABC,  Ac = \(\sqrt { { BC }^{ 2 }-{ AB }^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\quad cm\)
\(ar(\Delta ABC)=\frac { 1 }{ 2 } \times AB\times AC\)
\(=\frac { 1 }{ 2 } \times 3\times 4=6\quad { cm }^{ 2 }\)
ar(\(\Delta\)DBC) = ar(\(\Delta\)ABC) = 6 cm²
(Triangles on the same base and betweeen the same parallels)

ΔPTS and ΔPXS lie on same base and between the same parallels XT and PS
ar (PTS) = ar (PXS) ...(i)
Also, ΔSQR and ΔSQY lie on same base SQ and between the same parallels SQ and RY
ar (SQR) = ar (SQY) ...(ii)
Adding (i) & (ii), we get,
ar (PTS) + ar (SQR) = ar (PXS) + ar (SQY)
...(iii)
Adding ar (PQS) on both sides of (3), we get
ar (PTS) + ar (PQS) + ar (SQR)
= ar (PXS) + ar (PQS) + ar (SQY)
i.e., ar (PQRST) = ar (SXY)
Hence proved.

(i) According to the questions, the eqiuation of line is 
x+y=0..(i)
The line passes through a point whose y-coordinate is -19.5
y=-19.5
Put the value of y in equation (i), we get
x-19.5=0
x=19.5
So the required coordinates are (19.5-19.5)
(ii) According to the question, the equation of line is
x=y...(ii)
The line passes through a point whose x-coordinate is 20.5
x=20.5
Put this value of x in (i), we get
y=20.5
So the required coordinates are(20.5,20.5)
(iii) The coordinates of all the points lying on a line satisfy the equation of the line.
(iv) Obedience and respect for elders or parents.

(i) Steps of Construction
1. Draw the base BC = 5 em.
2. At point B make an angle XBC = 45°.
3. CutthelinesegmentBD=AB-AC(=2.8 cm) from the ray BX.
4. Join DC.
5. Draw the perpendicular bisector, say PQ of DC.
6. Let it intersect BX at a point A.
7. Join AC.
Then, ABC is the required triangle.

(ii) By measurement, AB = 13cm
(iii) By measurement, AC = 10.2cm
(iv) AB - AC = 13 - 10 .2 = 2.8cm
(v) Yes! Hari is true as by measurement ㄥACB = 112°.
The value 'wise' is depicted by comment of Hari.

Given: ABCD is trapezium in which AD=BC
To prove: Opposite angles of ABCD are supplementary.
Construction: Draw BE || AD
Proof: In quadrilateral ABED,
AB || DE  I Given
AD || BE  | By construction
∴ Quadrilateral ABED is a parallelogram.

| A quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.
∴ \(\angle BAD=\angle BED\)  ..........(1)
I Opposite angles of a parallelogram are equal
But  AD=BC   | Given
∴ BE=BC
∴ \(\angle BEC=\angle BCE\) | Angles opposite to equal sides of a triangle are equal
∴ \(\angle BEC=\angle BED=180°\) | Linear pair axiom
 ​​​​​​​\(\Rightarrow \angle BCE+\angle BED=180°\)   | From (2)
 \(\Rightarrow \angle BCE+\angle BAD=180°\)   | From (1)
\(\Rightarrow \angle BCD+\angle BAD=180°\)
\(\Rightarrow \) Opposite angles of ABCD are supplementary.

Given: In figure, EF is a line passing through the centre 0 of a circle. EF bisects chords AB and CD of the circle.
To Prove: AB || CD.
Proof: ∵ EF bisects chord AB
∴ OL bisects chord AB
∴ \(\angle OLB=\angle OLA=90°\)                        ...(1)
| ∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord
∵ From (1) and (2),
  \(\angle OLB=\angle OMC=90°\)
But these angles form a pair of equal alternate interior angles
∴ AB || CD.

Given: ABC is an equilateral triangle whose medians are AD, BE and CF.
To prove: AD = BE = CF
Proof: In \(\triangle ADC\) and \(\triangle BEC\)

AC = BC
\(\angle ACD=\angle BCE\)
AD is a median DC = DB = 1/2 BC
BE is a median EA = EC = 1/2 AC
AC = BC
DC = EC
\(\triangle ADC\cong \triangle BEC\) | SAS congruence rule
AD = BE  | C.P.C.T
Similarly, we can prove that
BF = CF ... (2)
CF = AD ....(3)
From (1), (2) and (3)
AD = BE = CF