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Solve the following fractional quadratic equation 14/(x^2-9)+1/(3-x)+(4-x)/(x+3)=7/(x+3), assuming x=/=+-3.

The first step is to find a common denominator. Since x^2-9 can be expanded to (x-3)(x+3) and 1/(3-x) can be written as -1/(x-3), we can see that all the terms contain either (x-3), (x+3), or both. Theref...

EH
Answered by Evelin H. Maths tutor
8615 Views

Solve x^2+4x-5=0 by completing the square.

x^2+4x-5=0(x+2)^2-(2)^2-5=0subtract 2^2 as (x+2)^2 = x^2+4x+4 when we want x^2+4x(x+2)^2-4-5=0(x+2)^2-9=0(x+2)^2=9square root both sidesx+2=+/-3x=-2+/-3x=-2+3 or x=-2-3x=1 or x=-5

EM
Answered by Evie M. Maths tutor
10674 Views

The radius of a pizza is 12cm, calculate the area of the pizza in terms of pi

To calculate the answer to this question we need to know the formula for the area of the circle. The area of a circle is:

A = Pi x r^2

The value of r is given in the question; r = 12. We the...

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Answered by Beth B. Maths tutor
9462 Views

Find the area under the curve y = (4x^3) + (9x^2) - 2x + 7 between x=0 and x=2

To be able to solve this, the equation needs to be integrated. To do this, simple integration needs to be applied where the power of 'x' increases by 1 and the coefficient of 'x' is divided by the new pow...

YS
Answered by Yash S. Maths tutor
5241 Views

GCSE 2011: Solve the simultaneous equations: y^2 = 2x + 29 and y = x - 3

This was the final question in AQA's 2011 non-calculator paper, and most top candidates will be looking to score all 5 marks that are on offer. Our usual technique for simultaneous equations isn't going t...

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Answered by Ross G. Maths tutor
8245 Views

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