2/4 = 0.5
3/4 = 0.75
7/8 = 0.875
3/8 = 0.375
Answer:
a) 42°F < x < 176°F
b) The inequality graph is attached.
c) No, This is because at 20° F benzene would not be in liquid form but in solid form.
Step-by-step explanation:
According to the Question,
a) For the benzene to remain in liquid form, the temperature of benzene must be less than the boiling point and greater than the boiling point. Let x be the temperature of benzene, For benzene to remain as liquid, its temperature must be between:
42°F < x < 176°F
b) The inequality graph is attached. The graph shows that the temperature of benzene must be between 42°F and 176°F so that it would be a liquid. The Closed circles represent that it is greater than 2.
c) No, This is because at 20° F benzene would not be in liquid form but in solid form. Because the temperature cannot go below 42 before it freezes, she would not have been able to conduct her research .
Answer:

Step-by-step explanation:
![\ln \dfrac{4y^5}{x^2}\\\\=\ln(4y^5) - \ln(x^2)~~~~~~~~~~~;\left[ \log_b\left( \dfrac mn \right) = \log_b m - \llog_b n \right]\\\\=\ln 4 + \ln y^5 - 2\ln x~~~~~~~~~~~~;[\log_b m^n = n \log_b m ~\text{and}~\log_b(mn) = \log_b m + \log_b n ]\\\\=\ln 4 + 5 \ln y -2 \ln x\\\\=\ln 4 -2 \ln x +5 \ln y](https://tex.z-dn.net/?f=%5Cln%20%5Cdfrac%7B4y%5E5%7D%7Bx%5E2%7D%5C%5C%5C%5C%3D%5Cln%284y%5E5%29%20-%20%5Cln%28x%5E2%29~~~~~~~~~~~%3B%5Cleft%5B%20%5Clog_b%5Cleft%28%20%5Cdfrac%20mn%20%5Cright%29%20%20%3D%20%5Clog_b%20m%20-%20%5Cllog_b%20n%20%5Cright%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%20%5Cln%20y%5E5%20-%202%5Cln%20x~~~~~~~~~~~~%3B%5B%5Clog_b%20m%5En%20%3D%20n%20%5Clog_b%20m%20~%5Ctext%7Band%7D~%5Clog_b%28mn%29%20%3D%20%5Clog_b%20m%20%2B%20%5Clog_b%20n%20%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%205%20%5Cln%20y%20-2%20%5Cln%20x%5C%5C%5C%5C%3D%5Cln%204%20-2%20%5Cln%20x%20%2B5%20%5Cln%20y)
Answer:
The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft
Step-by-step explanation:
Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²
For maximum volume, the side length, s of the cube must all be equal ;
Therefore area of one side = s²
Number of sides in a cube with top open = 5 sides
Area of surface = 5 × s² = 180
Therefore s² = 180/5 = 36
s² = 36
s = √36 = 6 ft
Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.