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R. S Jokers 1 year, 2 months ago

The total number of orange :- 360 And these oranges were distributed in among certain boys Three boys less each 4 more then orange The equation is :- 4x+3=360 4x=360-3 4x=357 x=357/4 x=89.25 So the students are 89.25

Pragya Kala 1 year, 3 months ago

No of boys=x
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Parbhuram. Bamniya 1 year, 4 months ago

Exercise 3.4
  • 2 answers

Manas Pandey 1 year, 3 months ago

Question 2

Kalimabid Khan 1 year, 4 months ago

Answer no 15 is cut
  • 1 answers

Manas Pandey 1 year, 3 months ago

Question 17
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R. S Jokers 1 year, 2 months ago

Then the first termis 2 then the 9th term is 26 Find the common difference We know that :- An=a+(n-1).d n=1 A1= a+0=2-------(1) Then n=9 A9=a+8d=26--------(2) Then By Elimination method Subtract eq(1) from (2) a+8d=26 a+0=2 8d=24 d=24/8 d=3 Puring the value of d in eqaction (2) a+8(3)=26 a+24=26 a=26-24 a=2 So the a first value is 2 And the d common difference is 3
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R. S Jokers 1 year, 2 months ago

Nonsense

R. S Jokers 1 year, 2 months ago

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  • 4 answers

R. S Jokers 1 year, 2 months ago

TanQ=P/H TanQ=1/cosQ TanQ=SinQ/cos Q

Pratham Ojha 1 year, 4 months ago

Tan theeta = p/b (perpendicular/base)

Sanjana Sajanasehwag 1 year, 7 months ago

Hmm

Jaat Sahb 1 year, 7 months ago

Sin/cos
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Preeti Dabral 1 year, 8 months ago

Given: {tex}\triangle {/tex}ABC {tex} \sim {/tex}{tex}\triangle {/tex}DEF
AM is a median in
{tex}\triangle {/tex}ABC and DN is the corresponding median in {tex}\triangle {/tex}DEF
To prove:
{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
Proof: {tex}\triangle {/tex}DBC {tex} \sim {/tex}{tex}\triangle {/tex}DEF
{tex}\Rightarrow {/tex} {tex}\angle{/tex} A = {tex}\angle{/tex} D, {tex}\angle{/tex} B = {tex}\angle{/tex} E, {tex}\angle{/tex} C = {tex}\angle{/tex} F and

{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}{/tex}
Also, {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}}{/tex}={tex}\frac{{A{B^2}}}{{D{E^2}}} = \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}{/tex}........(i)[area theorem]
Now,{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{\frac{1}{2}BC}}{{\frac{1}{2}EF}} = \frac{{BM}}{{EN}}{/tex}..............(ii)
In {tex}\triangle {/tex}ABM and DEN
{tex}\angle{/tex} B= {tex}\angle{/tex} E and {tex}\frac{{AB}}{{DE}} = \frac{{BM}}{{EN}}{/tex} [From (ii)]
{tex}\Rightarrow {/tex} {tex}\triangle {/tex}ABM {tex} \sim {/tex}{tex}\triangle {/tex}DEN [SAS similarity]
{tex}\Rightarrow {/tex}{tex}\frac{{area\vartriangle ABM}}{{area\vartriangle DEN}} = \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}} = \frac{{B{M^2}}}{{E{N^2}}}{/tex}............(iii)
From (i) and (iii), we get
{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}

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Pratibha Singh 1 year, 1 month ago

2

R. S Jokers 1 year, 2 months ago

Bro using an=a+(n-1).d

Sanjana Sajanasehwag 1 year, 7 months ago

You're so nonsense

Omprakash Choudhary 1 year, 9 months ago

A=2 d=4 An=a+(n-1)*d A=2

Anna Bhai 1 year, 9 months ago

2
  • 4 answers

Kittu Jaat 1 year, 8 months ago

77/2 cm square

Vicky Singh 1 year, 9 months ago

44

Kavita Dhariwal 1 year, 9 months ago

44

Anna Bhai 1 year, 9 months ago

44
  • 1 answers

Kittu Jaat 1 year, 8 months ago

X=-3 and -1/2
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Yuvraj Singh Deora 1 year, 10 months ago

Sorry I didn't take answer 😃

Preeti Dabral 1 year, 11 months ago

Incomplete question

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Dhananjay Tomar 2 years, 2 months ago

What is sector
  • 1 answers

Heena Kosar 2 years, 2 months ago

1 chapter
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Antika Gaur 2 years, 3 months ago

Exercise 2.3

Priyanka Rathore 2 years, 3 months ago

25.67

No Name No Name 2 years, 3 months ago

Pi r square theta upon m 360 degree

Queen Tnuu 2 years, 4 months ago

..
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  • 1 answers

Priyanka Rathore 2 years, 3 months ago

0

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