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 answer is 11 The answer is 11 The answer is 11
Answer:
ASA
Step-by-step explanation:
You can show the angles at either end of segment BC in triangles MBC and LCB are congruent, so you have two angles and the segment between. The appropriate theorem in such a case is ASA.
<O = 16 degrees vertical angles are equal
<O + 4x+90 = 180 DC is a straight line
16 +4x +90 = 180
106 +4x = 180
4x =74
x =18.5
16 +90 +2y= 180 DC is a straight line
106+2y=180
2y = 74
y= 37
Answer : y=37
x=18.5
<u><em>780 students</em></u>
<u><em></em></u>
<em>Set up a proportion.</em>
<em />
<em>75 / 125 = x/1300</em>
<em />
<em>125 10.4 = 1300</em>
<em />
<em>75 * 10.4 = 780</em>
<em />
<em>x=780</em>
