CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Line Frequency Graph: Line diagrams consist in drawing vertical lines, length of each line being equal to frequency. The values of variate x are presented on a suitable scale along x-axis. Line diagrams facilitate comparison but they are not attractive to see.
Histograms These graphs are used to graph grouped data. It is one of the most popular and commonly used devices for charting continuous frequency distribution. It consists of erecting a series of adjacent vertical rectangles on the section of X-axis with bases of equal width of the corresponding class intervals and the heights are so taken that the area of the rectangle are equal to the frequency of the corresponding classes.
Frequency Polygons A frequency polygon is formed by marking the mid-point at the top of horizontal bars and then joining these dots by a series of straight lines. The frequency polygons are formed as a closed figure with the horizontal axis, therefore a series of straight lines are drawn from the mid-point of the top base of the first and the last rectangles to the mid-point falling on the horizontal axis of the next outlaying interval with zero frequency.
Frequency Curve It is described as a smooth frequency polygon. A frequency curve is described in terms of its (i) symmetry (skewness) and (ii) degree of peakedness (kurtosis).Two frequency distributions can also be compared by superimposing two or more frequency curves provided the width of their class intervals and the total number of frequencies are equal for the given distributions. Even if the distributions to be compared differ in terms of total frequencies, they still can be compared by drawing per cent frequency curves where the vertical axis measures the per cent class frequencies and not the absolute frequencies.
Time Series Graph: When information is arranged over a period of time, it is called time series graph. In it, time (hour, day/date, week, month, year, etc.) is plotted along x-axis and the value of the variable along y-axis. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph (time series graph).

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Q-1 What are corrosponding angles ?

Posted by Aditya Juyal 5 years, 4 months ago

CBSE > Class 09 > Mathematics

2 answers

Reyansh Agarwal 5 years, 4 months ago

The angles which occupy the same relative position at each intersection.Where a straight line crosses two others.If the two lines are parallel,the corresponding angles are equal.

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Hitansh Garg 5 years, 4 months ago

Angels which are on either side on transversal of two parallel lines which keave one angle between them.

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ABC is an isosceles triangle in AB=AC.AD bisects exterior angle PAC and CD||AB. show that angle DAC=angle BCA and ABCD is a ||gm.

Posted by Zeenat Khanam 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 5 years, 4 months ago

{tex}\Delta{/tex}ABD where AB = AC
AD bisects {tex}\angle{/tex}PAC,
& CD {tex}\|{/tex} AB
To prove : {tex}\angle{/tex}DAC = {tex}\angle{/tex}BCA
Proof:
AD bisects {tex}\angle{/tex}PAC
Hence {tex}\angle{/tex}PAD = {tex}\angle{/tex}DAC = {tex}\frac{1}{2}{/tex} {tex}\angle{/tex}PAC ...(i)
Also, given
AB = AC
{tex}\therefore{/tex} {tex}\angle{/tex}BCA = {tex}\angle{/tex}ABC (Angles opposite to equal sides are equal) ... (ii)
For {tex}\Delta{/tex}ABC,
{tex}\angle{/tex}PAC is an exterior angle
so, {tex}\angle \mathrm{PAC}=\angle \mathrm{ABC}+\angle \mathrm{BCA}{/tex} (Exterior angle is sum of interior opposite angles)
{tex}\angle \mathrm{PAC}=\angle \mathrm{BCA}+\angle \mathrm{BCA}{/tex} (From (2) : {tex}\angle \mathrm{ABC}=\angle \mathrm{BCA}{/tex})
{tex}\angle \mathrm{PAC}=2 \angle \mathrm{BCA}{/tex}
{tex}\frac{1}{2} \angle \mathrm{PAC}=\angle \mathrm{BCA}{/tex}
{tex}\angle \mathrm{BCA}=\frac{1}{2} \angle \mathrm{PAC}{/tex}
{tex}\angle B C A=\angle D A C \quad\left(\text { From }(1): \angle D A C=\frac{1}{2} \angle P A C\right){/tex}
Hence proved

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Complete angle

Posted by Rehan Zaheer 5 years, 4 months ago

CBSE > Class 09 > Mathematics

2 answers

Reyansh Agarwal 5 years, 4 months ago

An angle whose measure is equal to 360 degree is called a complete angle.

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Namrata Jindal 5 years, 4 months ago

Angle of 360° is called complete angle

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Do char wali sankhya

Posted by Rehan Khan 5 years, 4 months ago

CBSE > Class 09 > Mathematics

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If the complement of a angle is one third of it's supplement find the angle

Posted by Kavya Dekivadiya 5 years, 4 months ago

CBSE > Class 09 > Mathematics

3 answers

Adithya Adi 5 years, 4 months ago

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Kanaka Kollati 5 years, 4 months ago

540

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Chiranjeet Kumar Arya 5 years, 4 months ago

45 Degree

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If two angles are complements of each other Then each angle is

Posted by Atul Yadav 5 years, 4 months ago

CBSE > Class 09 > Mathematics

4 answers

Kanaka Kollati 5 years, 4 months ago

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Sohil Sohil 5 years, 4 months ago

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Vedant Mohan 5 years, 4 months ago

The angle complement to itself is 45 degrees you can prove it by 90 degrees by 2 = 45 degrees and for the check method add 45 degree to 45 degree to get the 90 degree

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Sakshi Agrawal 5 years, 4 months ago

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Heron s formula

Posted by Nilam Nilam 5 years, 4 months ago

CBSE > Class 09 > Mathematics

2 answers

Ayush Mishra 5 years, 4 months ago

√s(s(s-a)(s-b)(s-c)

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Vedant Mohan 5 years, 4 months ago

Heron's formula is given in pg no 199 below the information of Heron

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1×2+2-5×3+4-5÷1000×100000÷10+5+5+5+5+3+4+3

Posted by Anirudh Vats 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Chiranjeet Kumar Arya 5 years, 4 months ago

-27

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If x+y=2 -x+y=5 Find x and y

Posted by Pranjal Salvi 5 years, 4 months ago

CBSE > Class 09 > Mathematics

2 answers

Kanaka Kollati 5 years, 4 months ago

1.X is 3 ,Y is 5=2 2.-X is 8 y is 3

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Yogita Ingle 5 years, 4 months ago

x+y=2 ......... (i)
-x+y=5 ............ (ii)
Adding (i) and (ii)
x + y -x + y = 2 + 5
2y = 7
y = 7/2
put y= 7/2 in (i), we get
x + 7/2 = 2
x = 2 - 7/2

x = -3/2

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How much electron in bulb

Posted by Shubham Yadav 5 years, 4 months ago

CBSE > Class 09 > Mathematics

0 answers

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The x coordinate is also called as

Posted by Vaibhav Pawar 5 years, 4 months ago

CBSE > Class 09 > Mathematics

4 answers

Kanaka Kollati 5 years, 4 months ago

Absissa

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Reyansh Agarwal 5 years, 4 months ago

Absissa

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Priyanshu Singh 5 years, 4 months ago

Absissa

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Rudra Dalal 5 years, 4 months ago

Absissa

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Homework help

Posted by Laxmi Kumari 4 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Harshit Gourlariya 5 years, 4 months ago

In which chapter

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Prove that vertically opposite angles are equal

Posted by Lalan Jha 5 years, 4 months ago

CBSE > Class 09 > Mathematics

0 answers

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How to solve algebra questions

Posted by Laxman Yadav 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sumit Singh 5 years, 4 months ago

The algebric Q we easily solve by using identities

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Prove that the sum of all the angle around a point is 360

Posted by Adarsh Kumar 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Laxman Yadav 5 years, 4 months ago

x+x+x+=360

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x^3+y^3+z^3=3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx

Posted by Rishabh Raoso Pawar 5 years, 4 months ago

CBSE > Class 09 > Mathematics

0 answers

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(x + y)^2 =x^2+2xy+y^2

Posted by Rishabh Raoso Pawar 5 years, 4 months ago

CBSE > Class 09 > Mathematics

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Ifan is 16 more than its complement then find its measure

Posted by Ꮆ ꫝᗰᎥᮘᎶ• 『乙ΞᗫꫝI』 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 5 years, 4 months ago

Let 1st complementry angle be x
Another= x + 16
x + (x + 16) = 90
2x + 16 = 90
2x = 90 - 16
x= {tex}\frac{74}{2}{/tex}= 37
Another angle= 37 + 16 = 53

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Find the supplement of 4/3 of right angle

Posted by Ꮆ ꫝᗰᎥᮘᎶ• 『乙ΞᗫꫝI』 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 5 years, 4 months ago

<div class="ans_text">{tex}\frac{4}{3}{/tex} of right angle
= {tex}\frac{4}{3}\times90{/tex}
= 120
Supplementary of 120
= 120 + x = 180
x = 180 - 120
x = 60

</div>

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in figure ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 centimetre if 2 root 5 find the area of the rectangle

Posted by Karan Singh 5 years, 4 months ago

CBSE > Class 09 > Mathematics

2 answers

Sia ? 5 years, 4 months ago

We have, AC = 10 cm and AD = 2{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} AC² = AD² + CD²
{tex}\Rightarrow{/tex} CD = {tex}\sqrt{AC^2 - AD^2}{/tex} = {tex}\sqrt{(10)^{2}-(2 \sqrt{5})^{2}}{/tex} = 4{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} ar(rect ABCE) = AD {tex}\times{/tex} CD = 2{tex}\sqrt{5}{/tex} {tex}\times{/tex} 4{tex}\sqrt{5}{/tex} cm²
= 8 {tex}\times{/tex} 5 cm² = 40 cm²