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A circle with center C has equation x^2 + y^2 + 8x - 12y = 12

Circle equation = (x - a)2 + (y - b)2 = r2Where Centre coordinates (a, b) and radius 'r'Therefore x2 + y2 + 8x - 12y = 12 is to be rewritten in this ...

Answered by Henry K. Maths tutor
8162 Views

Expand and simplify (x − 4)(2x + 3y)^2

Rewrite (x − 4)(2x + 3y)2 as (x - 4)(2x + 3y)(2x + 3y)
Multiply out the first two brackets to give -(2x2 + 3xy - 8x2 - 12y)(2x + 3y)
Multiply o...

Answered by Henry K. Maths tutor
14287 Views

Find the values of x and y when 3x+4y=18 and 4x+2y=14

Label equations: (a) 3x+4y=18(b) 4x+2y=14We need to be able to cancel out either the x's or the y's. The simplest choice is to double (b) so that we can cancel out the y's.2 x (b) : 8x+4y=28 (Give this a ...

Answered by Rosie E. Maths tutor
5465 Views

Solve the simultaneous equations 3x + 2y = 12 and 10y = 7x + 16

First we need to rearrange both equations into the form ax + by = c (where a, b and c are the integers). We can label these 1. and 2.We then have:3x + 2y = 12 1.7x -10y = -16 2.In order to find the x valu...

Answered by Anjali A. Maths tutor
3683 Views

Rewrite (2+(12)^(1/2))/(2+3^(1/2)) in the form a+b((c)^(1/2))

To solve:Rewrite the top part of the fraction to get it in terms of 3^(1/2)Rationalise the denominatorSimplify
The final answer is 2(3^(1/2)) - 2

Answered by Max M. Maths tutor
2924 Views

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