D=dollars earned
T=time spent painting the fence.
(1,5) (2,10) (3,15)
D=5T
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: 33 1/3%
Step-by-step explanation:
Percentage increase of the shirt will be calculated as:
= Price Increase / Old price × 100%
Price Increase = $20 - $15 = $5
Old price = $15
Percent increase = Price Increase / Old price × 100%
= 5/15 × 100
= 1/3 × 100
= 33 1/3%
900,000+80,000+500+7 is 980,507
The options aren't given, however, the range of amount spent could be calculated
Answer:
Kindly check explanation
Step-by-step explanation:
Paula purchase 3 dvd and 7 cds and spent between 90 and 100. Each dvd costs the same amount the price of cd is 10 select all amounts that could be the price of the dvd
Given that :
DVD costs the same amount
Number of DVD's = 3
Cost of cds = 10
Number of cds = 7
Total cost of cd's = 10 * 7 = 70
Let the price of DVD = x
Total amount spent = 90 - 100
Lowest price of DVD :
(amount spent - price of cd) / number of dvd
(90 - 70) / 3 = 20 /3 = 6.667
Highest price :
(100 - 70) / 3 = 30 / 3 = 10
Hence the pice of each DVD will range between /
6.6666 and 10.
Hence, prices listed within the range above are possible.
