Beats me!
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Well if the width is w then the length is 2w
You have to make an equation for the perimeter and solve it.
I don't understand this problems.
You don't necessarily have 15 on the first one do you?
If you have x on the first one then you have 60-x on all the other 5 combined.
I'd start by squaring both sides and solving it as a quadratic. Hope that helps.
Look here
http://web2.0calc.com/questions/how-do-i-type-in-fractions
In the sticky topics on the right hand side of your page is a thread called
"How do I type in fractions" Look in there.
NO
4(n-4)-1 = 4*n - 4*4 -1 = 4n-16-1 = 4n-17
If you have not heard of complex or imaginary numbers yet then there are no (real) solutions to this.
Google "scientific notation" abd then work through the "mathsisfun" site that it is bound to throw up.
$$\\w=\frac{-1}{-1/2cos(\theta^2)+(7/16)}\\\\ =-1\div \left(\frac{-1}{2cos(\theta^2)}+\frac{7}{16}\right)\\\\ =-1\div \left(\frac{-8}{16cos(\theta^2)}+\frac{7cos(\theta^2)}{16cos(\theta^2)}\right)\\\\ =-1\div \left(\frac{7cos(\theta^2)-8}{16cos(\theta^2)}\right)\\\\ =-1\times \left(\frac{16cos(\theta^2)}{7cos(\theta^2)-8}\right)\\\\ =\frac{16cos(\theta^2)}{8-7cos(\theta^2)}\\\\$$
Is that simple enough?
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