Alan

Factor the 2^x from each term on the left-hand side:

2^x(x² - 1) = 0

The only ways the left-hand side can be 0 is if x = 1 or x = -1 or x = -∞

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Add 3x to both sides

1 ≤ 7 + 3x

Subtract 7 from both sides

-6 ≤ 3x

Divide both sides by 3

-2 ≤ x

This says x must be greater than or equal to -2, so it can also be written as x ≥ -2

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If you have a quadratic equation like a*x² + b*x + c = 0 where a, b and c are constants, then the quadratic formula is x = (-b ± √(b² - 4*a*c))/(2*a)

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Consider the function y = x² 

This doesn't have an inverse (in the real number domain) because there isn't a unique value of x for a given y, but by writing x = y² before doing anything else you imply there is (to qualify as a function an expression must map an input to a unique output).

However, as you say, the important thing is being careful about the restrictions.  As long as you keep this in mind I will not say you can't do it Melody!  

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Yes, but in this case of course it is a simple linear function defined over all the real numbers for both x and y, so it works out ok either way.

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Here's a detailed breakdown of the steps needed to turn 2015 decimal into binary:

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The curved part of the perimeter of a semicircle is (1/2)*pi*diameter. To this you need to add the length of the diameter, so perimeter of semicircle = (1/2)pi*diameter + diameter.  

You can fill in the numbers.

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Really?  That's a new one on me!  (Forgive the pun!)

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The square root of -2 doesn't exist in the domain of real numbers, but it does exist in the domain of complex numbers.  It is i√2, where i is √(-1).

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You don't specify the value of the radius in the question civonamzuk, but if r = 1 then heureka's answer evaluates to the same as the one you give.

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