Four teams get to the semi final of the football association cup????
if weight of the feed, concentrate and tailing are F, C and T respectively and their corresponding assays f, c and t, then: F=C+T Ff=Cc+Tt Ff=Cc+t(F-C) therefore F/C= (c-t)/(f-t) which is the ratio of concentration. the plant recovery equals (Cc/Ff)*100 or based on the ratio of concentration the recovery is: Recovery=100*((f-t)c)/((c-t)f)
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As we know that the relation between T(F) and T(C) is given by: T(F)=9/5*T(C)+32 Now we want to know the point at which T(F) and T(C) both agrees i.e T(F)=T(C) So the above expression becomes, T(F)=9/5*T(F)+32 By solving above we get -40oC
The four temperature scales are Celsius ( C ) , Fahrenheit ( F ), Kelvin ( K ), and Rankine ( R ).T in C = ( 5/9 ) ( T in F - 32.0 )T in F = ( 9/5 ) ( T in C ) + 32.0T in K = T in C + 273.2T in R = T in F + 459.7
The temperature T in degrees Celsius (°C) is equal to the temperature T in degrees Fahrenheit (°F) minus 32, times 5/9: T(°C) = (T(°F) - 32) × 5/9 Convert 68 degrees Fahrenheit to degrees Celsius: T(°C) = (68°F - 32) × 5/9 = 20 °C
4 basic ones Celsius (C) Fahrenheit (F) Kelvin (K) Rankine (R)
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A-R-T-I-F-A-C-T
1. B 2. A 3. C 4. D American School Of Corr. Biology Examination 8, question 5. The snake has a tongue which is known as the jacobson's organ. Through this the snake smells the air in the environment and finds its prey in the surroundings. 6. t 7. t 8. t 9. f 10. f 11. t 12. t 13. t 14. t 15. t 16. t 17. t 18. f 19. f 20. f 21 d 22 c 23 c 24 a 25.c 26 c 27 c 28 b 29 d 30 b 31 d 32 b 33 a 34 d
1. B 2. A 3. C 4. D American School Of Corr. Biology Examination 8, question 5. The snake has a tongue which is known as the jacobson's organ. Through this the snake smells the air in the environment and finds its prey in the surroundings. 6. t 7. t 8. t 9. f 10. f 11. t 12. t 13. t 14. t 15. t 16. t 17. t 18. f 19. f 20. f 21 d 22 c 23 c 24 a 25.c 26 c 27 c 28 b 29 d 30 b 31 d 32 b 33 a 34 d