Question 325772
<font face="Garamond" size="+2">


Please read the instructions on the page where you submit your questions.  It is rather clear that you should only ask one question per submission.


1.  You are on the right track with the idea of a triangle.  But the height of the triangle is the DIFFERENCE of the heights of the ends of the cable.  35 minus 25 is 10.  So you have a right triangle with legs of 10 and 24.  Use Pythagoras to calculate the hypotenuse.  (check your work by comparing your answer to a 5-12-13 Pythagorean triple)


2.  The area of the first square is 3 times 3 is 9.  The area of the other is 4x times 4x is *[tex \LARGE 16x^2].  The sum is *[tex \LARGE 16x^2\ +\ 9].


*[tex \LARGE \left(4x\ +\ 3\right)^2\ =\ 16x^2\ +\ 24x\ +\ 9] is the area of a square that measures *[tex \LARGE 4x\ +\ 3] on a side and which is exactly *[tex \LARGE 24x] square units larger than the sum of the areas of your original two squares.


A couple of other things:


x^ does NOT represent *[tex \LARGE x^2].  The caret mark means raise the thing that precedes the caret mark to the power of the thing that comes after it.  If you want to say *[tex \LARGE x^2], write x^2.  If you want to say *[tex \LARGE x^3], write x^3, and so on.


*[tex \LARGE 4x\ \times\ 4x\ =\ (4x)^2\ =\ 16x^2\ \neq 4x^2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>