Answer:
Y = 3x^x is a graph that has exponential growth while y = 3^-x has exponential decay.
Y = 3x^x (-∞, 0) and (∞, ∞).
Y = 3x^-x (-∞, ∞) and (∞, 0).
Step-by-step explanation:
The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".
When graph 1 had positive ∞ for its x value, its y value was reaching towards positive ∞. When its x was reaching for negative ∞, its y was going for 0.
For graph 2, however, when its x was reaching for positive ∞, its x was reaching for 0. When its x was reaching for negative ∞, its y was going for positive ∞.
Here's an image of the graphs:
Answer:
c
Step-by-step explanation:
Here's how this works:
Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:
![[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?](https://tex.z-dn.net/?f=%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%2B%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%20%3D%3F)
In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

Put everything over the common denominator now:

Since
, we will make that substitution:

We could separate that fraction into 2:
×
and 
Therefore, the simplification is
sec(x)csc(x)
EQUIVALENT WOULD BE THE ANSWER