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  • 2 answers

Khwahis Parveen 1 year, 5 months ago

x+y=14 …(i) x-y=4 …(ii) x=4+y …(iii) Now substituting the value of x in equation (i) x+y=14 4+y+y=14 4+2y=14 2y=14-4 2y=10 y=5 Now put the value of y in equation (iii) x=4+y x=4+5 x=9 Answer:- the value of y=5 and x=9

Harsh Kumar 1 year, 5 months ago

x = 14 - y Put ↑ in x-y = 4 ... 14 -y-y = 4 → 14 -2y = 4 → -2y = 4-14 → - 2y = -10 → y = -10/-2 Therefore,y = 5 And x = 9
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  • 2 answers

Abhishek Meena Ganj Basoda 1 year, 5 months ago

Zgh

Khwahis Parveen 1 year, 5 months ago

I think you have written the question wrong.
  • 1 answers

Saksham Kumar 1 year, 5 months ago

Where
  • 1 answers

Apeksha Kumari 1 year, 5 months ago

125
  • 1 answers

Gaurav Choudhary 1 year, 5 months ago

E+h=EH
  • 1 answers

Gaurav Choudhary 1 year, 5 months ago

4
  • 1 answers

Licy Soren 1 year, 5 months ago

0
  • 2 answers

Beena Manoj 1 year, 5 months ago

LCM 3360

Beena Manoj 1 year, 5 months ago

HCF =96
  • 2 answers

Arpan Uniyal 1 year, 5 months ago

-12/35

Shivam Yadav 1 year, 5 months ago

-12÷35 answer (-12/35)
  • 1 answers

Swaran Singh 1 year, 5 months ago

Let assume that √2 is an rational number and √2/1 = a/b , where a and b are integers and co-prime , b ≠0 . b√2 = a By squaring both sides, we get 2b²= a² _ (1) Here, a² is divisible by 2 and a also divisible by 2. Now , let a=2c , where c is an integer . By squaring both sides, we get a²= 4 c² By Substituting it in eq ( 1) 2b²= 4c² b² = 2c² Here , b² is divisible by 2 , also b is divisible by 2. Therefore, 2 is a common factor of a and b . This contradicts the fact that a and b are not co - prime. Therefore , our assumption is wrong.
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  • 1 answers

Khwahis Parveen 1 year, 5 months ago

Firstly, 615-5= 610 963-7= 956 find the HCF of 610 and 956 then you get the answer

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