\(\frac { 1 }{ { 17 }^{ 3 } } \)

\(5+2\sqrt { 3 } -3\sqrt { 2 } \)

0,  \(\frac{1}{27}\)

Let x be the weight of Ram and Ravi each. On gaining 2 kg, weight of Ram and Ravi will be (x+2) kg each. According to Euclid's second axiom, when equals are added to equals, the wholes are equal. So, weights of Ram and Ravi are again equal.

\((i)\frac { 1 }{ 2 } =\frac { 1\times 5 }{ 2\times 5 } =\frac { 5 }{ 10 } =0.5\ |q=2\)
\(\\ (ii)\frac { 1 }{ 4 } =\frac { 1\times 25 }{ 4\times 25 } =\frac { 25 }{ 100 } =0.25\ |q={ 2 }^{ 2 }\)
\(\\ (iii)\frac { 3 }{ 8 } =\frac { 3\times 125 }{ 8\times 125 } =\frac { 375 }{ 1000 } =0.375\ |q={ 2 }^{ 3 }\)
\(\\ (iv)\frac { 31 }{ 25 } =\frac { 31\times 4 }{ 25\times 4 } =\frac { 124 }{ 100 } =1.24\ |q={ 5 }^{ 2 }\)
\(\\ (v)\frac { 7 }{ 125 } =\frac { 7\times 8 }{ 125\times 8 } =\frac { 56 }{ 1000 } =0.056\ |q=5^{ 3 }\)
\(\\ (vi)\frac { 13 }{ 20 } =\frac { 13\times 5 }{ 20\times 5 } =\frac { 65 }{ 100 } =0.65\ |q={ 2 }^{ 2 }\times 5\)
\(\\ (vii)\frac { 29 }{ 16 } =\frac { 29\times 625 }{ 16\times 625 } =\frac { 18125 }{ 10000 } =1.8125 |q={ 2 }^{ 4 }\)
The property that q must satisfy is that the prime factorization of q must have only powers of 2 or powers of 5 or both.

\(\therefore \ p(0)=(0-1)(0+1)=(-1)(1)=-1\)
\(\\ p(1)=(1-1)(1+1)=(0)(2)=0\)
and \(p(2)=(2-1)(2+1)=(1)(3)=3\)

\(2x-\sqrt{7}+0=0;\ a=2,\ b=-\sqrt{7}.,\ c=0\) Yes.

480 cm²

Slant height (l)         = 21 m
Diameter of base     = 24 m
\(\therefore \) Radius of base (r) = \(\frac { 24 }{ 2 } m=12m\)
\(\therefore \) Total curved surface area of the cone
\(=\pi r\left( l+r \right) \)
\(\\ =\frac { 22 }{ 7 } \times 12\times \left( 21+12 \right) \)
\(\\ =\frac { 22 }{ 7 } \times 12\times 33=\frac { 8712 }{ 7 } =1244.57{ m }^{ 2 }.\)

(i) Pob. of less than 41
\(=\frac{8+12}{90}=\frac{20}{90}=\frac{2}{9}\)
(ii) Prob.of more than 50 \(=\frac{20+13+17+5}{90}=\frac{55}{90}=\frac{11}{18}\)
(iii)Prob. of marks between 41 and 80
\(=\frac{15+20+13+17}{90}=\frac{65}{90}=\frac{13}{18}\)

Volume of the shot-put=\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 35 }{ 10 } \times \frac { 35 }{ 10 } \times \frac { 35 }{ 10 } \)
\(=\frac { 11\times 49 }{ 3 } { cm }^{ 3 }\)
Mass of the shot-put\(=\frac { 11 }{ 3 } \times 49\times \frac { 78 }{ 10 } \)
= 1401.10 gm.

2.x + y = 8
⇒ y=8-2x
Table of solution

We plot the points (4, 0) and (0, 8) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 2.x + y = 8.

From graph, we see that this line intersects the x-axis at the point (4, 0) and the y-axis at the point (0,8).

AB || CD
and a transversal EF intersects them
\(\therefore \)\(\angle EGB=\angle GHD\) 
| corresponding Angles
\(\Rightarrow 2\angle EGM=2\angle GHL\)
GM and HL are the bisectors of \(\Rightarrow \angle EGM=\angle GHL\) respectively
\(\Rightarrow \angle EGM=\angle GHL\)
But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF.
\(\therefore \) GM || HL.

Steps of Construction
1. Draw a line segment MN = 5 cm.
2. With M as centre and 5 cm as radius, draw an arc on one side of MN.
3. With N as centre and 5 cm as radius, draw another arc on the same side of MN to intersect the former arc at L.
4. Join LM and LN. Then, \(\Delta \) LMN is the required equilateral triangle.

5. Taking M as centre and any radius, draw an arc to intersect the line segments MN and ML at P and Q respectively.
6. Next, taking P and Q as centres and with 1 the radius more than \(\frac { 1 }{ 2 } \) PQ, draw arcs to intersect each other, say at R.
7. Draw the ray MR. This ray MR is the required bisector of the \(\angle M\).

Mean
\(=\frac {76+44+45+87+71+72+82+83+41+32+75+32+46+78+17+70+84+12+77+74} {20}\)
\(=\frac {1198}{20}=59.9\)

Quotient = x³-3x² + x-2
Remainder = 2
So, -2 should be added in p(x).
So that it is completely divisible by q(x)

In trapezium ABCD,
AB II DC and AD = BC
Through C, draw
CE II DA.
DC II AE and CE is transversal.

∠1=∠2
∠3=∠1
∠2=∠3=∠1
∠2+∠3=2∠1
∠A+∠C=∠3+∠2+∠4
=2∠1+∠4
Also ∠1=∠5
∠A+∠C=∠1+∠4+∠5=180°
Similarly, we can show that ㄥB+ㄥD=180°
Hence, the opposite angles of an isosceles trapezium are supplementary.

\(\angle\)ACD = 90°, \(\angle\)AED = 90°   (angle in semi-circle)
\(\angle\)ADE = 180° - (60° + 90°)
= 30° (angle sum property)
\(\angle\)ABC = 180° - \(\angle\)CDA
= 180°- 70° = 110°  (ADCB is a cyclic quadrilateral)
\(\angle\)BCA = 180 - (110° + 30°) = 40°  (Angle sum property)
\(\angle\)BCD = \(\angle\)BCA + \(\angle\)ACD
Therefore
\(\angle\)BCD = 40° + 90° = 130°.