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Prove algebraically that 0.256565656... can be written as 127/495

Let x = 0.2565656...Then 10x = 2.565656.... and 1000x = 256.565656...So 1000x - 10x = 256.565656... - 2.565656... = 254990x = 254 x = 254/990 = 127/495

Answered by Oliver T. Maths tutor
13096 Views

Solve the simultaneous equations algebraically.

x^2 + y^2 = 29y - x = 3Rearrange the second equation such that one variable, either x or y, is the subject. I will rearrange to make x the subject. x = y - 3. Substitute the new equation into equation 1 i...

Answered by Jane A. Maths tutor
2193 Views

Prove that (3n+1)²-(3n-1)² is a multiple of 4 taking into account that n is a positive integer value

  1. Square the brackets (3n+1)²= (3n+1)(3n+1) = 9n²+3n+3n+1 = 9n²+6n+1 (3n-1)²= (3n-1)(3n-1) = 9n²-3n-3n+1 = 9n²-6n+12. Write out the full equation (9n²+6n+1) - (9n²-6n+1) = 9n²+6n+1-9n²+6n-1 = 12n3. Ex...
Answered by Nalin K. Maths tutor
5640 Views

How do you solve two simultaneous equations? (i.e. 5x + y =21 and x - 3y =9)

There are two ways of solving the simultaneous equations. The easiest one is to write two equations beneath each other and then try to get either x or y values the same on both of them by multiplying the ...

Answered by Paulina M. Maths tutor
2408 Views

Find the value (8/125)^-2/3

(8/125)^-2/3We can write this as ((8/125)^1/3)^-2So cube rooting we then get (2/5)^-2Then if we get the inverse we get (5/2)^2And finally we square the answer to get 25/4

Answered by Nabeel N. Maths tutor
6684 Views

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