Answer: 7.8 ⋅ 
Step-by-step explanation:
To write a linear expression in standard form, rearrange the terms in alphabetical order.
7.8 ⋅ 
The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?
we have that
e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1
then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10
the answer is x=1.10
The slope (m) = -13.
We can use the slope formula:
m = (y2 - y1) / (x2 - x1)
Let (x1, y1) = (7, 26)
(x2, y2) = (12, -39)
Plug in the values into the formula:
m = (y2 - y1) / (x2 - x1)
m = (-39 - 26) / (12 - 7)
m = -65/5
m = -13
Answer:
Part 1) The expression for the perimeter is
or 
Part 2) The perimeter when z = 15 ft. is 
Step-by-step explanation:
Part 1)
we have

Find the roots of the quadratic equation
Equate the equation to zero

Complete the square
Group terms that contain the same variable, and move the constant to the opposite side of the equation

Factor the leading coefficient


Complete the square. Remember to balance the equation by adding the same constants to each side


Rewrite as perfect squares

-----> root with multiplicity 2
so
The area is equal to
![A=625(z-0.12)(z-0.12)=[25(z-0.12)][25(z-0.12)]=(25z-3)^{2}](https://tex.z-dn.net/?f=A%3D625%28z-0.12%29%28z-0.12%29%3D%5B25%28z-0.12%29%5D%5B25%28z-0.12%29%5D%3D%2825z-3%29%5E%7B2%7D)
The length side of the square is 
therefore
The perimeter is equal to



Part 2) Find the perimeter when z = 15 ft.
we have

substitute the value of z

Answer:
They should be at each exabit for about 51 min
Step-by-step explanation: