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.
The equation is equivalent to x2/25 + y2/(25/<span>9)</span>= 1, it is an ellipse equation
the center is C(0,0)
a=5; b= 5/3, so a>b it is a horizontal major axis type (the x-axis)
all that you do is to find all characteristics by following all required formula
You’d set up a proportion. I did 25/15=7x-40/18.
Then you would cross multiply to get 15(7x-40)=450
Distribute the 15.
105x-600=450
Add 600 to both sides, so the -600 cancels out.
Then you’re left with 105x=1050
Divide both sides by 105 and you get x=10
Answer:
GCF is 3
Step-by-step explanation:
Given the expression -9x + 15y
First, we need to find ther individual factors
-9x = -3 *3 * x
15y = 3 * 5 * y
Since 3 is common to both factors
−9x+15y = 3(-3x+5y)
Hence the GCF is 3