Answer:
h= 0.25c +4
Step-by-step explanation:
➀ Define variables
Assuming that you are asking for the equation representing the height of the stack, let's start by letting the height of the stack be h inches and the number of cups be c.
➁ Find the increase in height with every additional cup
If an additional cup is stacked on the first cup, the height increment is 0.25 inches. When 2 cups are added, the height increment is 2(0.25)= 0.5 inches. Thus the expression for the increase in height is 0.25c, where c is the number of cups as we have already defined in step 1.
➂ Find the total height of the stack
Since the height of the first cup remains constant at 4 inches tall, the total height of the stack can be represented by the equation:
h= 0.25c +4
Given:
In a two-digit number, the tens digit is 5 less than the units digit.
The number itself is five more than three times the sum of its digits.
To find:
The number.
Solution:
Let the two digit number is ab. So,

Tens digit is 5 less than the units digit.
...(i)
The number itself is five more than three times the sum of its digits.



...(ii)
Using (i) and (ii), we get





Putting b=8 in (i), we get


Therefore, the required number is 38.
the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.
from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.
the standing up sides are simply rectangles of 8x3.
if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D5%5C%5C%20p%3D24%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%285%29%2824%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bjust%20for%20one%20octagon%7D%7D%7BA%3D60%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwo%20octagon%27s%20area%7D%7D%7B2%2860%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Beight%20rectangle%27s%20area%7D%7D%7B8%283%5Ccdot%208%29%7D%5Cimplies%20120%2B192%5Cimplies%20312)
The correct answer is 0.73737373 or answer C. An irrational number is a number that cannot be expressed as a fraction for any integers and. . Irrational numbers have decimal expansions that neither terminate nor become periodic
It has to be either A or B