2.27 x 10⁻⁴ = 0.000227

(7 x 10²) (5.2 x 10⁷) = 36.4 = 10⁹ = 3.64 x 10¹⁰

\(\left( \cfrac { { 3 }^{ 5 } }{ { 3 }^{ 8 } } \right) ^{ 5 }\times \left( { 3 }^{ -5 } \right) =\left( { 3 }^{ 5-8 } \right) ^{ 5 }\times { 3 }^{ -5 }=\left( { 3 }^{ -3 } \right) ^{ 5 }\times { 3 }^{ -5 }={ 3 }^{ -3\times 5 }\times { 3 }^{ -15 }={ 3 }^{ -15-5 }=\cfrac { 1 }{ { 3 }^{ 20 } } \)

\(\\ \cfrac { { 3 }^{ 2 } }{ { 3 }^{ -2 } } ={ 3 }^{ 2 }\times { 3 }^{ 2 }=9\times 9=81\)

Find the midpoint of the class intervals and tabulate it.

The points are A(5,0) B(15,4) C(25,6) D(35,8) E(45,12) F(55,10) G(65,14) H(75,5) I(85,7) J(95,0).
In the graph sheet, mark the midpoints along the x- axis and the frequency along the y- axis.
We take the imagined class as 0 – 10 at the beginning and 90 – 100 at the end , each with frequency ‘zero’.
From the table, plot the points. We draw the line segments AB, BC, CD, DE, EF, FG, GH, HI, IJ to obtain the required frequency polygon ABCDEFGHIJ.

Mark the class intervals along the x-axis and the number of students along the y-axis . Draw a histogram for the given data and mark the midpoints of the rectangles and join them by lines. We get frequency polygon. Note that the first and last edges of the frequency polygon meet at the mid points of the left and right vertical edges of first and last rectangles. Because imagined class intervals do not exist in the marks (refer the above note).

Represent the pocket money along x- axis and number of students along the y–axis.
Draw a histogram for the given data. Now, mark the midpoints of the upper sides of the consecutive rectangles. Also mark the midpoints of two imagined class intervals 0-10 and 90-100 whose frequency is 0 on x- axis. Now, join all the midpoints with the help of ruler. We get a frequency polygon imposed on the histogram.

1. The histogram represents the collection of weights from std VIII.
2. There are maximum 9 students in 30-35 kg weight.
3. There are 7(= 2 + 5) students who are under weight
4. There are 3 students who are obese
5.  There are 16( = 9 + 7) students in the 30-40 kg weight group.

The given distribution is discontinuous. If we represent the given data as it is by a graph we shall get a bar graph, as there will be gaps in between the classes. So, convert this into a continuous distribution using the adjustment factor 0.5.
The first class interval can be written as 9.5-15.5 and the remaining class intervals are changed in the same way. There are no changes in frequencies.
The new continuous frequency table is

The histogram is constructed as below

The given data is a continuous frequency distribution. The class intervals are drawn on x axis and their respective frequencies on y-axis. Classes (ages) and its frequencies (number of people) are taken together to form a rectangle.
The histogram is constructed as given below

Given: side = 4 cm

Steps:
(i) Draw a line segment LA = 4 cm.
(ii) At L, construct LX ⊥LA.
(iii) With L as centre, draw an arc of radius 4 cm and let it cut LX at P.
(iv) With A and P as centres, draw arcs of radius 4 cm each and let them cut at M.
(v) Join AM and PM. LAMP is the required square
Calculation of area:
Area of square LAMP = a² sq.units
= 4 x 4 = 16sq.cm

Given: BE = 5 cm and BN = 3 cm

Steps:
(i) Draw a line segment BE = 5 cm.
(ii) At B, construct BX ⊥BE .
(iii) With B as centre, draw an arc of radius 3 cm and let it cut BX at N.
(iv) With E and N as centres, draw arcs of radii 3 cm and 5 cm respectively and let them cut at A.
(v) Join EA and NA.
(vi) BEAN is the required rectangle
Calculation of area:
Area of rectangle BEAN = l × b sq.units
= 5 x 3 = 15sq.cm

Given: NS = 9 cm and ET = 8 cm


Steps
(i) Draw a line segment NS = 9 cm.
(ii) Draw the perpendicular bisector XY to NS. Let it cut NS at O.
(iii) With O as centre, draw arcs of radius 4 cm on either side of
O which cut OX at T and OY at E.
(iv) Join NE, ES, ST and TN.
(v) NEST is the required rhombus
Calculation of area:
Area of rhombus NEST=\(\cfrac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }sq.units\) 
=\(\cfrac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }sq.units\) 
= \(\cfrac { 1 }{ 2 } \times 9\times 8=36sq.cm\)

Given: RO = 5 cm and RS = 8 cm


Steps:
(i) Draw a line segment RO = 5 cm.
(ii) With R and O as centres, draw arcs of radii 8 cm and 5 cm respectively and let them cut at S.
(iii) Join RS and OS.
(iv) With R and S as centres, draw arcs of radius 5 cm each and let them cut at E.
(v) Join RE and SE.
(vi) ROSE is the required rhombus.
Calculation of area:
Area of rhombus ROSE =\(\cfrac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }\) sq.units
= \(\cfrac { 1 }{ 2 } \times 8\times 6=24sq.cm\)

(-7)^{x + 2} x(-7)⁵ = (-7)¹⁰
(-7)^{x + 2 + 5} = (-7)¹⁰
Since the bases are equal, we equate the exponents and get
x + 7 = 10
x = 10 - 7 = 3