Question:
John read the first 114 pages of a novel, which was 3 pages less than ⅓ of the novel. Write an equation to determine the total number of pages (P)
Answer:
114 = ⅓P - 3
Explanation:
Given
Number of pages read = 114
Total pages in novel = p
The relationship between the pages read by John and the total pages is analysed as follows:
3 less than ⅓ of total pages means:
⅓ of total pages - 3
Recall that P represents the total pages in the novel
So, the expression becomes
⅓ * P - 3
⅓P - 3
This means that the pages read by John is ⅓P - 3
This implies that the equation to determine the number of pages in the novel is
⅓P - 3 = 114
Solving further to get the actual number of pages;
Multiply both sides by 3
3(⅓P - 3) = 114 * 3
3 * ⅓P - 3 * 3 = 114 * 3
P - 9 = 342
Add 9 to both sides
P - 9 + 9 = 342 + 9
P = 351
Hence the number of pages is 351
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
Answer: 12.4 feet
Explanation:
If there is a smooth transition and there is no change in slopes, energy considerations can be used
The cube has a kinetic energy of
ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when there is a gain in potential energy
pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
If there is no loss in energies,
pe = ke
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
h = 2000 /322 = 6.211 (ft)
= h / sin(30) = 12.4 ft
Answer:
Rubber-like solids with elastic properties are called elastomers.
Explanation:
I’m sorry what’s the question?