Odd functions are those that satisfy the condition
f(-x)=-f(x)
For example, check if x^3 is odd =>
f(x)=x^3
f(-x) = (-x)^3
-f(x)=-x^3
Since (-x)^3=-x^3, we see that f(x)=x^3 is an odd function.
In fact, polynomials which contain odd-powered terms only are odd. (constant is even)
As an exercise, you can verify that sin(x) is odd, cos(x) is even.
On graphs, odd functions are those that resemble a 180 degree rotation.
Check with graphs of above examples.
So we identify the first graph (f(x)=-x^3) is odd (we can identify a 180 degree rotation)
Odd functions have a property that the sum of individually odd functions is
also odd. For example, x+x^3-6x^5 is odd, so is x+sin(x).
For the next graph, f(x)=|x+2| is not odd (nor even) because if we rotate one part of the graph, it does not coincide with another part of the graph, so it is not odd.
For the last graph, f(x)=3cos(x), it is not odd, again because if we rotate about the origin by 180 degrees, we get a different graph. However, it is an even function because it is symmetrical about the y-axis.
Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component 
. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.
However, let's prove in a more formal way that
![\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%3A%5C%20%5Cmathbb%7BR%7D%20%5Cto%20G%2C%5Cquad%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is an isomorphism.
First of all, it is injective: suppose 
. Then, you trivially have 
, because they are two different matrices:
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%2C%5Cquad%20%5Cphi%28y%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
Secondly, it is trivially surjective: the matrix
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is clearly the image of the real number x.
Finally, 
and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have
![\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)](https://tex.z-dn.net/?f=%20%5Cphi%20%28x%2By%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%2By%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cphi%28x%29%20%5Ccdot%20%5Cphi%28y%29)
Answer:
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Step-by-step explanation:
The problem consists in finding the new length of a running trail knowing the original length and the extension required. So the new length will be equal to original length + extension. New length = 2.826 miles + 1.46 miles = 2.826 miles + 1. 460 miles (I added a 0 in the place of the thousandths for the second summand to make clear the sumation) = 4.286 miles. Then<span> the answer is 4.286 miles.</span>
That would mean we're doubling the number of peanuts to get raisins. So, we would have 2x+y for the total amount.
