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someone show me step by step how to solve these two questions
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nebbercracker
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So these are two questions on a past paper I am stuck with. One is inverse proportion and the other is simultaneous equation. Please could someone explain step by step how to solve them ?
Inverse proportion
h is inversely proportional to the square of r
when r =5 , h=3.4
find the value of h when r=8
simultaneous equation
x(squared) + y(squared) = 9
x + y = 2
help with these questions will be much appreciated
Inverse proportion
h is inversely proportional to the square of r
when r =5 , h=3.4
find the value of h when r=8
simultaneous equation
x(squared) + y(squared) = 9
x + y = 2
help with these questions will be much appreciated
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rede121
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M.C. Math
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(Original post by nebbercracker)
So these are two questions on a past paper I am stuck with. One is inverse proportion and the other is simultaneous equation. Please could someone explain step by step how to solve them ?
Inverse proportion
h is inversely proportional to the square of r
when r =5 , h=3.4
find the value of h when r=8
simultaneous equation
x(squared) + y(squared) = 9
x + y = 2
help with these questions will be much appreciated
So these are two questions on a past paper I am stuck with. One is inverse proportion and the other is simultaneous equation. Please could someone explain step by step how to solve them ?
Inverse proportion
h is inversely proportional to the square of r
when r =5 , h=3.4
find the value of h when r=8
simultaneous equation
x(squared) + y(squared) = 9
x + y = 2
help with these questions will be much appreciated
- Take the second equation: y=2-x. Substitute for y in the first equation.
x2 + (2-x)2 = 9
- Expand the brackets
x2+x2-4x+4=9
2x2-4x-5=0. Solve for x (i'd use the quadratic formula as it doesn't factorise, or if you prefer to 'complete the square', divide through by 2), then substitute each value of x into the second equation to find y.
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