Answer: The correct conclusion is(B) The functions f(x) and g(x) are reflections over the y-axis.
Step-by-step explanation: Two functions f(x) and g(x) are given as follows:

We know that if f(-x) = g(x), then the functions are reflections over Y-axis and if - f(x) = g(x), then the functions are reflections over X-axis.
We have,

So, the function g(x) is a reflection of f(x) over Y-axis.
The graph of f(x) and g(x) are drawn in the attached file. From there, it is clear that the functions are reflections over Y-axis, not reflections over X-axis.
So, options (A) is incorrect and option (B) is correct.
From the table, we have

So, as the value of 'x' increases, the value of f(x) increases and value of y(x) decreases.
Therefore, f(x) is an increasing function and g(x) is a decreasing function. So, option (C) is incorrect.
Also, we have

So, both the functions have same initial value. So, option (D) is also incorrect.
Thus, the correct conclusion is (B) The functions f(x) and g(x) are reflections over the y-axis.
The optimum pinhole diameter for a camera box with a length of 10 centimeters is 0.446 mm
<h3><u>
Solution:</u></h3>
Given that,
<h3><u>The optimum diameter d (in millimeters) of the pinhole in a pinhole camera can be modeled by:</u></h3>
![d = 1.9[(5.5 \times 10^{-4})l]^{\frac{1}{2}}](https://tex.z-dn.net/?f=d%20%3D%201.9%5B%285.5%20%5Ctimes%2010%5E%7B-4%7D%29l%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where l is the length (in millimeters) of the camera box
<h3><u>Find the optimum pinhole diameter for a camera box with a length of 10 centimeters</u></h3>
l = 10 cm
We know that,
10 cm = 100 mm
<em><u>Therefore, plug in l = 100 in given formula</u></em>
![d = 1.9[(5.5 \times 10^{-4}) \times 100]^{\frac{1}{2}}\\\\d = 1.9[5.5 \times 10^{-4} \times 10^2]^{\frac{1}{2}}\\\\d = 1.9[5.5 \times 10^{-2}]^{\frac{1}{2}}\\\\d = 1.9 \times 5.5^{\frac{1}{2} \times 10^{-1}}\\\\d = 0.19 \times 2.345207\\\\d = 0.4455 \approx 0.446](https://tex.z-dn.net/?f=d%20%3D%201.9%5B%285.5%20%5Ctimes%2010%5E%7B-4%7D%29%20%5Ctimes%20100%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cd%20%3D%201.9%5B5.5%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%2010%5E2%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cd%20%3D%201.9%5B5.5%20%5Ctimes%2010%5E%7B-2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cd%20%3D%201.9%20%5Ctimes%205.5%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5E%7B-1%7D%7D%5C%5C%5C%5Cd%20%3D%200.19%20%5Ctimes%202.345207%5C%5C%5C%5Cd%20%3D%200.4455%20%5Capprox%200.446)
Thus the optimum pinhole diameter for a camera box with a length of 10 centimeters is 0.446 mm
8 times four is less than 6 squared by 4, because 8*4 = 32 and 6^2 = 36 then 36-32 = 4.
Answer:
Step-by-step explanation:
6/8 in simplest form is 3/4 but value is still the same so
1. no
Area = Length x Width
The Length is 2 inches
The Width is 5/6 inches
2 * (5/6) = 10/6
simplify and you get....
5/3