With a pie chart, we can compare parts to a whole.

Unorganised form of the data is raw data.

Concave polygon; given, figure is a closed curve with two angles more than 180°. So, another name of this shape is concave polygon.

The diagonals of the quadrilateral HOPE are HP and OE.

The prime factorisation of 529 is
529 = 23\(\times\)23
By pairing the prime factors, we get
529 = \(\underline { 23\times 23 } \)
So, \(\sqrt { 529 } \) = 23

\(\begin{matrix}\quad 4\quad A \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad C\quad B\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, B, and C whose values are to be found.
Studying the addition in the one's column, we have A + 8 and we get 3 from this, i.e. a number whose one's digit is 3, for this A has to be 5.
\(\because\) A + 8 = 5 + 8 = 13
Now, for sum in ten's column, we have
1 + 4 + 9 = CB \(\Rightarrow\) 14 = CB
Here, B = 4 and C = 1
Therefore, the puzzle is solved as shown below:
\(\begin{matrix}\quad \quad 4\quad 5 \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 14\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 4 and C = 1

\(\begin{matrix}\quad 3\quad A \\ +\quad 2\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad B\quad 2 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in the one's column, we have A + 5 and we get 2 from this, i.e. a number whose one's digit is 2, for this A has to be 7.
\(\because\) A + 5 = 7 + 5 =12
Now, for sum in ten's column, we have
1 + 3 + 2 = B \(\Rightarrow\) B = 6
Therefore, the puzzle is solved as shown below:
\(\begin{matrix}\quad 3\quad 7 \\ +\quad 2\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 6\quad 2 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 7 and B = 6

\(9\over7\)

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

In class interval 100 - 150, the number of peeps is 11 i.e. maximum and in-class interval 50 - 100 the number of people is 2, i.e. minimum.

\(1\over4\)

We know that,
Central angle of a component = \((\frac{Component\quad value}{sum\quad of \quad the\quad component\quad values}\times 360)^o\)
Here, total amount = Rs 7200,
 Cntral angle for an item = \((\frac{Amount\quad spent\quad on\quad the\quad item}{Total\quad amount}\times 360)^o\)
The central angles of the sectors representing different items are computed in the following table:

Now, to construct the pie chart representing the above data, we follow the following steps:
Step I Draw a circle of an appropriate radius.
Step II Draw a vertical radius of the circle drawn in Step I.
Step III Choose the largest central angle. Here, largest central angle is of 200°. Draw a sector with central angle 200° in such a way that its one radius coincides with the radius drawn in Step II and another radius is in its clockwise direction.
Step IV Construct other sectors representing other items in clockwise sense in descending order of magnitudes of their central angles except the sector representing miscellaneous expenses. This sector is to be drawn in the last.
Step V Shade the sectors, so obtained by different designs and label them as shown in figure given below to obtain the required pie chart.

(i) Size of class interval = Upper-class limit - Lower class limit
= 125 - 100 = 25
(ii) The class 200-225 has the highest frequency i.e. 140.
(iii) The class 300-325 has the lowest frequency i.e. 20.
(iv) The upper limit of the class interval 250-275 is 275.
(v) The class interval 150 - 175 and 225 - 250 have the same frequency i.e. 55.

From the given bar graph, we find that,
i) The information given by the bar graph is the number of students in Class VIII in various academic years in the school.
(ii)In the year 2004-05, increase in the number of students is maximum.
(iii) The number of students is maximum in the year 2007 - 08.
(iv) False, because the number of students during 2005-06 is more than twice that of 2003-04.

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\(\frac{27}{8}\)

3^-2=\(\frac{1}{3^{2}}=\frac{1}{9}\)    [∵ \(a^{-m}=\frac{1}{a^{m}}\)]

we have, \(-13\over19\)
\(\therefore\)The multiplicative inverse of \(-13\over19\) is\(19\over-13\) 

We have, -13
\(\therefore\)The multiplicative inverse of -13 is \(-1\over13\)

We have, x =\(11\over15\)
LHS = -(-x) = -\(({11\over15})=​​​​{11\over15}=X=RHS\)
S0, - (-x) = x is verified for x =\(11\over15\)

we have, \(19\over-6\) so, additive inverse of \(19\over-6\) is \(19\over6\)

we have, \({2\over-9}\) so, additive inverse of \({2\over-9}\) is \({2\over9}\)

we have,\(-6\over-5\)=\({6\over5}\) so, additive inverse of \(({-6\over-5})\) is \(-6\over5\)

= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) -\frac { 1 }{ 4 } +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } -\frac { 1 }{ 4 } \) (Using commutativity)
= \(\frac { 2 }{ 5 } \left[ \frac { -3 }{ 7 } +\frac { 1 }{ 14 } \right] -\frac { 1 }{ 4 } \) (using distributivity)
= \(\frac { 2 }{ 5 } \left[ \frac { -6+1 }{ 14 } \right] -\frac { 1 }{ 4 } =\frac { 2 }{ 5 } \times \frac { -5 }{ 14 } -\frac { 1 }{ 4 } \)
= \(\frac { -1 }{ 7 } -\frac { 1 }{ 4 } =\frac { -4-7 }{ 28 } =\frac { -11 }{ 28 } \)