+2446 There is a special formula that will aide us in this problem. It is the following:
\(K=\frac{1}{2}mv^2\)
K = the kinetic energy, in joules
m = the mass, in kilograms
v = velocity in meters per second
We have all that information already! Let's use it to find the mass of this cheetah.
| \(23610=\frac{1}{2}m(31)^2\) | Let's do some simplification. |
| \(23610=\frac{961}{2}m\) | Multiply the right hand side by \(\frac{2}{961}\) to obtain the mass, in kilograms. |
| \(m=\frac{23610*2}{961}\approx49.14kg\) | |
+2446 In order to simplify this expression, we must create a common denominator. Otherwise, subtracting complicated fractions is not feasible. Currently, \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) does not have this, so we must create one.
The denominator of the first term, \(a^2-1\) can be factored because it is a difference of two squares. Sometimes, factoring the denominator fully is helpful since it can make identifying the least common multiple all that more painless. In this case, \(a^2-1=(a+1)(a-1)\).
We now have \(\frac{a^2+1}{(a+1)(a-1)}-\frac{a-1}{a+1}\). Well, would you look at that! It is clear here that \((a+1)(a-1)\) is the least common multiple of the denominators since \(a+1\) is a factor of the product of binomials. Let's manipulate \(\frac{a-1}{a+1}\) so that we can create the desired common denominator.
| \(\frac{a-1}{a+1}\) | As aforementioned, let's convert this into a fraction with a common denominator. |
| \(\frac{a-1}{a+1}*\frac{a-1}{a-1}\) | Of course, the actual value of the fraction remains unchanged since I am really multiplying the fraction by one. Now, let's combine. |
| \(\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | |
Ok, we have now changed the problem from \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) to \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\). This transformation is indeed a positive one because we can now combine the fractions. We have now overcome the first stumbling block.
| \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | Combine the fractions now that the same denominator of both terms exist. |
| \(\frac{a^2+1-(a-1)(a-1)}{(a+1)(a-1)}\) | Let's determine the product of the binomials. Since both binomials are identical, we can utilize the rule that \((x-y)^2=x^2-2xy+y^2\) |
| \(\frac{a^2+1-(a^2-2a+1)}{(a+1)(a-1)}\) | Distribute the negative sign to all the terms it contains in parentheses. |
| \(\frac{a^2+1-a^2+2a-1}{(a+1)(a-1)}\) | Look at that! In the numerator, the a2-term will cancel out and so will the constants! |
| \(\frac{2a}{(a+1)(a-1)}\) | We can do the multiplication of the binomials in the denominator again. We already know what the product is. |
| \(\frac{2a}{a^2-1}\) | Of course, let's not forget the restrictions for the value of a. |
| \(\frac{2a}{a^2-1}, a\neq\pm1\) | If a was equal to the restricted values, a division by zero would occur, which is forbidden. |
