Answer:
For a function determining if it is a function or not is if they have two of the same x values it’s not a function if they don’t it is a function
Step-by-step explanation:
Answer: True
Step-by-step explanation:
We know that the quadratics polynomials have this form:

You need to remember that the degree of a polynomial is given by its highest exponent.
You can observe in the quadratic polynomial
that the highest exponent is "2", which is the exponent of the term 
Therefore, since the highest exponent is "2", you should conclude that the degree of the polynomial is 2.
Then, you can say that a quadradic polynomial has degree 2.
Features of a Parabola
The equation of a parabola can be written as:
Answer:
f'(x) > 0 on
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

To find its decreasing interval :

2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
Answer:
127.28 feet
Step-by-step explanation:
Given
Shape of basketball diamond = Square
Length of a side = 90 feet
If the length of a side is 90 feet then the length of other sides is 90 feet.
By finding how far the catcher, at home plate, throw to get the runner "out", we are basically finding the distance from 2nd base to home
This means we're calculating the diagonal of the field..
(Since the field is a square, it has a right angle at 1st base)
Let the diagonal be representing by x.
From Pythagoras,
x² = 90² + 90²
x² = 8100 + 8100
x² = 16200
x = √16200
x = 127.2792206135785
x = 127.28 feet ------ Approximated