8640

To plot the given points in each case, firstly we draw the X and Y-axes and then plot the given points.
On plotting the points P(1,1), Q(2,2), R(3, 3) and S(4, 4) on the graph paper and join them. Then, we get the following graph:

Yes, it is clear from the above graph that given points P, Q, R and S lie on a line

32

\({-13\over15},{-12\over15},{-11\over15},{-10\over15},{-9\over15}\)

We know that, there are 2n natural numbers lie between the squares of n and (n + 1).
To find natural numbers lie between 9² and 10².
Here, n = 9 and n + 1= 10 
\(\therefore\) Total natural numbers between 9² and 10²
= 2n = 2 x 9 = 18

2x(3x + 5xy) = 2x\(\times\)3x + 2x\(\times\)5xy
= (2\(\times\)3)\(\times\)x\(\times\)x + (2\(\times\)S)x\(\times\)xy
= 6\(\times\)x²+10 x²y = 6x²+10x²y

Each solid and its corresponding top front and side views are given below:

7x - 9 =16 \(\Rightarrow\) 7x =16 + 9  [transpoting 9 from LHS to RHS]
7x =25  \(\Rightarrow \frac { 7x }{ 7 } =\frac { 25 }{ 7 } \)
\(\Rightarrow\) \(x=\frac { 25 }{ 7 } \), Which is the required solution

(i) 3x\(\times\)5y = 15xy
(ii) 9y\(\times\)4y²= 36y3
(iii) 4ab\(\times\)5bc= 20ab²c
(iv) 2l²m x 3lm² = 6l²m³

(i) 37° C
(ii) 37.5° C
(iii) 11 am
(iv) 7 a.m. to 8 a.m.

The lowest observation = 1
The highest observation = 48
∴ We have

Central angle of the sector representing
(a) Hindi language = \({40\over72}\) x 360° = 40 x 5° = 200°
(b) English language =\(12\over72\) x 360° = 60°
(c) Marathi language =\(9\over72\) x 360° = 45°
(d) Tamil language = \(7\over72\) x 360° = 35°
(e) Bengali language = \(4\over72\)x 360° = 4 x 5 = 20°
Thus, the required pie chart is given below:

We have, 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 
\(\Rightarrow\) 15y - 60-2y + 18 + 5y + 30 = 0
\(\Rightarrow\)15Y + 5y - 2Y - 60 + 18 + 30 = 0
\(\Rightarrow\) 20Y - 2Y - 60 + 48 = 0
\(\Rightarrow\)18y - 12 = 0 \(\Rightarrow\) 18y =12  [transposing -12 from LHS to RHS]
\(\Rightarrow\)  \(y=\frac { 12 }{ 18 } \)
y = \(\frac { 2 }{ 3 } \) , which is the required solution

The prime factorisation of 5832
5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Therefore, \(\sqrt [ 3 ]{ 5832 } \)
= \(\sqrt [ 3 ]{ \underline { 2\times 2\times 2 } \underline { 3\times 3\times 3 } \underline { 3\times 3\times 3 } } \)
= 2 x 3 x 3
=18

Given number is 1764.
By using prime factorisation, we get
1764 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 7 \(\times\) 7
= (2\(\times\)3\(\times\)7)²
\(\therefore \ \sqrt { 1764 } \) = 2 \(\times\) 3 \(\times\) 7 = 42
Hence, the square root of 1764 is 42.

We have, \(\frac{25\times t^{-4}}{5^{-3}\times 10\times 10^{-8}}\)
=\(\frac{25\times(5)^{3}\times t^{8}}{10\times t^{4}} \quad [\because a^{-m}=\frac{1}{a^{m}}]\)
=\(\frac{25\times125\times t^{8-4}}{10} \quad [\because \frac{a^{m}}{a^{n}}=a^{m-n}]\)
=\(\frac{25\times125}{10}\times t^{4}\)
=\(\frac{5\times 125}{2}\times t^{4}=\frac{625}{2}t^{4}\)

We have
\({3\over7}+({-2\over21})\times({-5\over6})={3\over7}+[({-2\over21})\times({-5\over6})]\)
\( \\ ={3\over7}+({-5\over-63})={3\over7}+{5\over63}={27+5\over63}={32\over63}\)

2499

we have \(\frac { 2x+1 }{ 3x-2 } =\frac { 9 }{ 10 } \) 
\(\Rightarrow\) 10(2x + 1) = 9(3x - 2)       [by cross-multiplication]
\(\Rightarrow\) 20x +10 = 27x - 18  \(\Rightarrow\) 20x - 27x = - 18 - 10 [transposing 27x to LHS and 10 to RHS]
\(\Rightarrow\) -7x = - 28  \(\Rightarrow\) \(\frac { -7x }{ -7 } =\frac { -28 }{ -7 } \)  [dividing both sides by - 7]
\(\therefore\)  x = 4

Given, RUNS is a parallelogram and SU, RN intersect each other at O. Then, OS = 20 cm, OU = (y + 7) cm,
OR =16 cm, and ON = (x + y) cm
We know that, diagonals of a parallelogram bisect each other.

 ∴ OU = OS ⇒ y + 7 = 20
⇒ y = 20 - 7=  13 and ON = OR
⇒ x + y = 16 ⇒ x + 13=16    [∵ y=13 cm]
⇒ x = 16 - 13 = 3
Hence, the values of x and yare 3 cm and 13 cm, respectively.

We know that, the sum of three angles of a triangle is 180°.

Let the given triangle be ABC.
∴ ㄥA + ㄥB+ ㄥC =180°
⇒  30⁰ + 90⁰+ ㄥC=180° ⇒  ㄥC+120⁰=180°
⇒  ㄥC = 180° - 120° = 60°
Now, ㄥEBC + ㄥABC = 180° [by linear pair]
∴  x +90° = 180° ⇒ x = 180° - 90° = 90°
Similarly, ㄥFCA + ㄥACB = 180° [by linear pair]
⇒ y+60⁰=180° ⇒ y=180⁰-60⁰=120°
and ㄥDAB + ㄥBAC = 180° [by linear pair]
⇒ z + 30° = 180° ⇒ z = 180° - 30° = 150°
∴ x + y + z = 90° + 120° + 150° = 360°
Hence, the value of x + y + z is 360°

6 outcomes are 1, 2, 3, 4, 5 and 6.

Square; since, in a square all the sides and angles are equal.