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HELP WITH HOMEWORK NEED IT BY TODAY!!!!!!!!!!!!!!!!!!!!!?

OK MY TEACHER is really strict so try to work things out please and thank you!!!!!!! here's the question A FREIGHT TRAIN with 80 cars transports coal and iron ore. Each car carries either 100 tons of coal or 90 tons of iron ore. question A) Let c represent the number of cars carrying coal. Write and simplify an expression in terms of c for the total weight of the freight transported by the train. B) Suppose 28 of the trains cars carry coal. What is the total weight of all the freight? and the last question simplfy the expression a+(2a)+ (10-3a) THANKS REMEMBER WORK OUT THE PROBLEMS!!!!!!!!!!! MY TEACHER IS REALLY STRICT!!!!!!!!!! :)

Public Comments

1. well i suck at word problems but the last one all you do is combine like terms :) so the expression i'm pretty sure would be 6a-10 simplfied
2. let's solve B and work the info back into A a train has 80 cars cars carry coal or iron C+I=80 28 cars carry coal 28*100 tons=2800 tons I=80-C I=80-28 I=52 I=52*90=4680 tons total weight 7480 tons how to use that in A) C+I=80 C(100)+I(90)=weight I=80-C C(100)+(80-C)(90)= weight simplify a+(2a)+(10-3a) a+2a+10-3a the a's cancel out gives you 1+2a-3a= -10 0= -10 null solution [you can get that when you inadvertently divide by zero somewhere]
3. don't panic, you usually get answers here within minutes A) there are 80 cars in total c is the number of cars carrying coal so the number of cars carrying iron would be 80 - c now the weight carried by each coal car is 100 ton, and the weight carried by each iron car is 90 ton then, c coal cars carry 100c tons (80 - c) iron cars carry 90*(80 - c) tons so, total weight = 100c + 90(80 - c) = 100c + 7200 - 90c = 10c + 7200 B) 28 of the trains carry coal we can just put in c = 28 in the equation we obtained above total weight = 10*28 + 7200 =280 + 7200 = 7480 last question: a+(2a)+ (10-3a) =3a + 10 - 3a = 10 (3a and -3a cancel out)
4. For part A you need to set up two equations, one for the number of cars and one for the total weight. For the number of cars: 80 = c + i where c is the number of coal cars and i is the number of iron cars For the weight: W = 100c + 90i We can rearrange the first equation to get i = 80 - c and plug this into the second equation to get: W = 100c + 90(80-c) For part B we know what c is so we can just plug it into the equation we just go to solve for W. W = 100(28) + 90(80-28) = 7480 tons For the last you can just add all the terms with a in them together to get a+2a+10-3a = a+2a-3a+10 = 10