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  • 3 answers

_Jass_ Mahey_ 1 year, 7 months ago

Cos

Tribhuwan Kumar 1 year, 7 months ago

Keeta - theeta

Naani Kanumala 1 year, 7 months ago

3/root16
  • 2 answers

Abhishek Mukherjee 1 year, 7 months ago

Hi

Sarika Gupta 1 year, 7 months ago

HCF of 306 and 657 is 9. We know that: HCF *LCM = Product of the two numbers 9*LCM =306*657 LCM = 306*73 LCM = 22338. ✓
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Nibha Kumari 1 year, 7 months ago

Middle term splitting method.. x²-5x-6 X²-6x+x-6 x(x-6)+1(x-6) (x-6)(x+1)
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Faizal Khan 1 year, 3 months ago

Uuzyssd
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Nashid Khan 1 year, 7 months ago

45 ,25
  • 2 answers

Vibha Kumari 1 year, 7 months ago

6^n (2×3)^n 2^n×3^n Here , it do not contain 2^m× 5^n as a factor so 6^n does not end with the digit 0 Hope it helps 🙃

Not Fine Bro 🎭 1 year, 7 months ago

6×6..........n times
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Not Fine Bro 🎭 1 year, 7 months ago

12.33
  • 3 answers

Not Fine Bro 🎭 1 year, 7 months ago

Cost of all chairs - 2800 Cost of all tables - 2800

Sankalp Dinesh 1 year, 7 months ago

Answer - Let the cost of 1 chair be x and 1 table be y Therefore, 5x + 4y = ₹5600 -----(1) And, 4x + 3y = ₹4340 -----(2) 4 So, We use elimination method. Multiplying Equation (1) by 3 & Multiplying Equation (2) by 4 Then Equation (1) will become 15x + 12y = ₹16800 ----(3) And, Second Equation will become 16x + 12y = ₹17360 ---(4) Then, to eliminate y we will subtract Equation (4) from Equation (3) i.e. - x = - 560 Therefore, x = ₹560 ---(5) So, putting the value of Equation (5) in Equation (1) Then, 5x + 4y = ₹5600 5(560) + 4y = 5600 2800 + 4y = 5600 4y = 5600 - 2800 4y = 2800 y = 2800/4 y = ₹700 ------(6) By the Equation (5) & (6), We can say that the cost of a chair is ₹560 and the cost of a table is ₹700.

Manpreet Singh 1 year, 7 months ago

Let 1 chair be x. and let 1 table be y. According to question case1. 5x+4y=₹5600.(Equation1) Case2. 4x+3y=₹4340.(equation2) by 4 in equation 1 and byBy elimination method Multiply 5 in equation 2 and substract equation 1 and 2 , we get 5(4x+3y)=₹4340 20x+15y=₹
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Sankalp Dinesh 1 year, 7 months ago

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