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  • 2 answers

Goutam Dalal 7 years, 8 months ago

No

Hitesh Kumar 7 years, 8 months ago

Yes
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Goutam Dalal 7 years, 8 months ago

30

Dipti Sudha 7 years, 8 months ago

30
  • 2 answers

Kushal Saini 7 years, 8 months ago

((2)1/2)4+3^2+3-2^2+1/2+5(5)1/2 (2/4)4 +9+3-4+1/2+25+1/2 8/16+9+3-4+1/2+25+1/2 8/16+1/2+1/2+33 then take the LCM 32+8+8///4 +33 48/4 +33 12+33 =44

Kushal Saini 7 years, 8 months ago

45
  • 1 answers

Kshitij Sharma 7 years, 8 months ago

Beta it means to solve
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  • 1 answers

Shruti Kumari 7 years, 8 months ago

Yes , we can write 0as a form of rational number i.e.0/1
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  • 2 answers

Priyanshu Kumar 7 years, 8 months ago

17/4

Arshdeep Singh 7 years, 8 months ago

4.1,4.2,4.3,4.4,4.5
  • 2 answers

Saurav Kumar 7 years, 8 months ago

1152 is correct answer

Sahil Sagar 7 years, 8 months ago

288
  • 1 answers

Kajal Dhurve 7 years, 8 months ago

3/1*10/10=30/10 4/1*10/10=40/10 so the rational number between 3and 4are 31/10 , 32/10 , 33/10 ,34/10 ,35/10 ,36/10 .....39/10 take any six
  • 2 answers

Adarsh Uppal 7 years, 8 months ago

Yes

Seetha K 7 years, 8 months ago

Yes
  • 2 answers

Seetha K 7 years, 8 months ago

No, bcoz 1/2 is a rational number but not an integer

Sagar Paul 7 years, 8 months ago

False
M
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  • 3 answers

Shehdeep Singh 7 years, 8 months ago

2

Hitesh Nagpal 7 years, 8 months ago

2√2

Shruti Kumari 7 years, 8 months ago

2 root 2
  • 1 answers

Ayush Bansal 7 years, 8 months ago

1/2+2/5 = (5+4)/10 = 9/10
  • 1 answers

Shreya Roy 7 years, 8 months ago

Bcoz if we multiply (a+b )×(a+b)=a^2+ab+ba+b^2 so ab+ba=2ab .hence verified the above alzebric identity .hope it helps
  • 1 answers

Awantika Tiwari 7 years, 8 months ago

(2x)^3 + (1)^3 + 3×2x ×1(2x+1) =8x^3+1+6x(2x+1) =8x^3+1+12x^2+6x
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Armaanjot Singh 7 years, 8 months ago

kx³+9x²+4x-10=0 k(-3)³+9(-3)²+4(-3)-10=0 -3k+27-12-10=0 -3k+27-22=0 -3k+5=0 -3k=-5 k=-5/3

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