Answer:
The length of the side of an equilateral traingle
inches
Step-by-step explanation:
Given that the area of an equilateral triangle is given by

It can be written as
Square inches (1)
To find the length of the side s os an equilateral triangle
Given that area of an equilateral triangle is
square inches
It can be written as

square inches
It can be written as

square inches (2)
Now comparing equations (1) and (2) we get


Dividing by
on both sides we get




Therefore
inches
Therefore the length of the side of an equilateral traingle
inches
Answer: 
Step-by-step explanation:
Formula: 
Since the base is a circumference, we can replace B for the area of a circumference formula.

What we have here is 12 ft as the diameter of the circumference. A diameter is twice the radius, therefore we can conclude that if we take the diameter and divide it by 2, it will give us the radius.
this is the radius.
Now plug all this information into your formula.

Our values are in feet but the question requires cm. Let's convert from
to
.
Normally, our conversion factors are raised to the power of 1, but in this case it's raised to the power of 3. So, first let's see how many cm are 1 ft.

Here is where magic comes. We can raise both to the power of 3, in order to find the cubic ft and cm that we need as our conversion factors.

Now we have our conversion factors.

Answer:
The log-mean-temperature-difference is 24.03⁰C
Step-by-step explanation:
First we need to know if the heat exchanger is in parallel flow or counter-flow. However, counter flow arrangement is best used to recover heat.
L.M.T.D for counter flow is given as;
![L.M.T.D =\frac{(T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1)}{2.3log[\frac{T_h_f_1 -T_c_f_2}{T_h_f_2 -T_c_f_1}]}](https://tex.z-dn.net/?f=L.M.T.D%20%3D%5Cfrac%7B%28T_h_f_1%20-T_c_f_2%29-%28T_h_f_2%20-T_c_f_1%29%7D%7B2.3log%5B%5Cfrac%7BT_h_f_1%20-T_c_f_2%7D%7BT_h_f_2%20-T_c_f_1%7D%5D%7D)
where;
Thf₁ is the initial temperature of the hot fluid = 80°C
Tcf₂ is the final temperature of the cold fluid = 51.5°C
Thf₁ - Tcf₂ = 80 - 51.5 = 28.5⁰C
Thf₂ is the final temperature of the hot fluid = 30°C
Tcf₁ is the initial temperature of the cold fluid = 10°C
Thf₂ - Tcf₁ = 30 - 10 = 20⁰C
![L.M.T.D = \frac{28.5 -20}{2.3Log[\frac{28.5}{20}]} \\\\L.M.T.D = \frac{8.5}{0.3538} =24.03^oC](https://tex.z-dn.net/?f=L.M.T.D%20%3D%20%5Cfrac%7B28.5%20-20%7D%7B2.3Log%5B%5Cfrac%7B28.5%7D%7B20%7D%5D%7D%20%5C%5C%5C%5CL.M.T.D%20%3D%20%5Cfrac%7B8.5%7D%7B0.3538%7D%20%3D24.03%5EoC)
Therefore, the log-mean-temperature-difference is 24.03⁰C
122
286-42=144
144\2=122
122(peter)+164(malco)=286(total weight)