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Show algebraically that (4n-3)^2 - (2n+5)^2 is always a multiple of n-4

First we expand the brackets by squaring each side(4n-3) 2 = (4n-3) (4n-3)= 16n 2 - 24n + 9(2n+5) 2 = (2n+5) (2n+5)= 4n 2 + 20n + 25Remember the expression is (4n-3) 2 - (2n+5) 2 so we subtract the expanded ...
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Answered by Ella B. • Maths tutor
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Put the following in order of size, smallest first: 8/sqrt3, sqrt6*sqrt2, sqrt48-sqrt27

First part of the question is to recognise that these are surds and we will need to simplify them. Then it is asking to place the values from smallest to highest. In order to simplify the surds we have to fi...
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Answered by Sandeep S. • Maths tutor
4371 Views

solve the Simultaneous equations: y= x+6 and y=2x^23

2x 2 = x + 62x 2 - x - 6 = 0 [+-1 x-12]2x 2 - 4x + 3x - 6 =02x(x-2) 3(x-2) = 0(2x + 3)(x-2) = 0x =2 or -3/2 y = 2 + 6 y = 8 y = -3/2 + 6y = 9/2 (2,8)(-3/2, 9/2)
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Answered by Lehana D. • Maths tutor
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Prove that 0.5757... (recurring) = 19/33. Hence, write 0.3575757... (recurring) as a fraction in its lowest terms.

Two parts to the question. Let's focus on part one:Let x = 0.575757... (1)This means that 100x = 57.575757... (2)If you subtract (1) from (2), we get: 99x = 57Divide both sides by 99: x = 57/99Simplify: x = ...
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Answered by Oliver V. • Maths tutor
7992 Views

Find the positive solution to b^2 +5b – 6 =o

(b-1)(b+6)=0 b= 1 and -6 Positive Solution = 1
Answered by • Maths tutor
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