Answer: the area is 20 square centimeters.
Step-by-step explanation:
The area of an arc of angle x in a circle of radius r is:
A = (x/3.14)*3.14*r^2 = x*r^2
here we have x = (4/5)*3.14 and r = 5cm
Then the area is:
A = (4/5)*(5cm)^2 = 20cm^2
 
        
                    
             
        
        
        
Answer:
54 I think 
Step-by-step explanation:
I just did 15 divided by 9 and got 5/3 and then did 90 divided by 5/3 and got 54
 
        
             
        
        
        
I think its saying find the mixed numbers but keep the denominator
        
                    
             
        
        
        
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
 
        
                    
             
        
        
        
Answer:
Hope this is right.
1. 7(4x-5)
2. 28x-35
Step-by-step explanation: