-=RX-7 KID=-

write two equations in two variables. solve



1. Salt solutions of 60% and 15% are mixed to produce a 450 cm3 mixture of a 25% solution. How much of each solution is required?



2. flying with the wind, an airplane takes 2 h to fly 360 mi from amarillo to dallas. Flying against the wind, the sirplane takes 3h to fly the same distance. What is the speed of the wind and the speed of the airplane in calm air?



3. Mrs. Birch invested $1200, some at 6%, the rest at 4.5% per year. the retuen from the 4.5% investment exceeded that from the 6% investment by $12. How much was invested at each rate?



Thanks for any help guys.

Ghiadesigns

I think the answer to #2 is wind 30 mph and calm air travel is 150 mph

gsl-se addict

1. The total amount for both solutions is 450 cm3. So if we call the amount of the 60% solution, x, then the amount of the 15% solution, y, is equal to 450-x.



So,



y= (450 - x)



.60*x+.15*y=.25*450



--> .60*x+.15(450-x)=.25*450

--> .60*x-.15*x=.25*450-.15*450

--> .45*x=112.5-67.5=45

--> x=45/.45=100



So y= 450-x = 450-100 = 350



So, 100 cm3 of the 60% and 350 cm3 of the 15%



2. With the wind:

360 mi/ 2h = 180 mi/hr (ground speed)



Against the wind:

360 mi/3 hr = 120 mi/hr (ground speed)



180 mi/hr = plane speed + wind speed

120 mi/hr = plane speed - wind speed



P.S. = 180 mi/hr - W.S. (from first equation)



Insert into second equation



120 mi/hr = (180 - W.S.) - W.S.

-2*W.S. = (120-180)



W.S. = -(120-180)/2 = 30 mi/hr



Now, plug the wind speed into either one of the 2 equations. Using the first equation:



180 = P.S. +30 mi/hr



P.S. = 180 - 30 = 150 mi/hr



So the plane speed in calm air is 150 mi/hr and the wind speed during the trip was 30 mi/hr



3. For the interest problem say x dollars is at 6% and y dollars is at 4.5% and we know the total invested is $1200.



So: x+y=1200



We also know that the amount at 4.5% return $12 more dollars than the 6% investment (assuming this return was after a period of 1 year)



y*4.5%=x*6%+$12



if decimal form:



(1) y*0.045 = x*0.06 + $12



and



(2) x+y=$1200



so y=1200-x from equation (2)



Now insert into equation (1)



(1200-x)*0.045 = x*0.06 + $12

--> (1200-x)*0.045 -x*0.06= $12

--> -x*0.045 -x*0.06= $12-(1200*0.045)

--> -x(0.105)=12-(1200*0.045)=12-54

--> x=-(12-54)/0.105 = 400



Now put this back into equation (1):



x+y=1200



400+y=1200



y=800



So she had $400 at the 6% rate and $800 at the 4.5% rate.



Hope this helps.



Kent

Ghiadesigns

answer to #1 100cm3 of 60% and 350cm3 of 15%

you are on your own on the last one, I am going to bed

g night



ok someone beat me to it never mind

Dramon_Killer

...owned.

-=RX-7 KID=-

Hey guys thanks for all the help got one last one here too.







write a system of equations in 3 varavbilews for each problem and solve



the measure of the largest angle of a triangle is equal to twice the sum of the measures of the other two angles. the sum of the measures of the smallest and largest angles is 3 times the measure of the remaining angle. find the measures of the three angles.



Good luck no one can get this one... so far

gsl-se addict

Okay, call angles x, y, and z.



First, we know that the sum of all of the angles = 180 deg (from geometry)



(1) x+y+z=180



Let's say that z is the largest angle and x is the smallest, so:



(2) z=2*(y+x) - from the first statement in the problem



(3) x+z = 3*y - from the second problem statement



Now, we can plug equation (3) into equation (1) to get:



y+3*y = 180



--> 4*y=180

--> y = 180/4 = 45 deg



plug this info into eqn. (2) to get:



z=2*(45+x)

-->z=90+2*x



then this can be put into either eqn. (1) or (3). If we put it in eqn. (1), we get:

x+y+z=180

--> x+45+z=180

--> x+45+(90+2*x)=180

--> x+2*x+135=180

--> 3x=180-135

--> x=(180-135)/3 = 45/3 =15 deg



So, x=15 deg, y=45 deg, and that means z= (180-45-15) = 120 deg.



Kent

-=RX-7 KID=-

Kent your the ******* man, are u a math teacher?

gsl-se addict

No man. I am a graduate student in Nuclear Engineering. I am trying to finish up my PhD in spring term. I guess after 10 years of college, you get pretty good at solving problems like this. I key to problems like this is reading the problem carefully and figuring what information that you know and how to put it into equation form. From there it is just plugging one equation into another and solving for the variable. I am sure that you will get it after awhile and it will be no problem for you to do these problems.



Once the problems become larger, you will be looking at solving these equations in terms of matrices. It is the same concept as far as writing out the equations, but in matrix form you solve for all variables at once instead of plugging on equation into another. You can probably imagine what a time saver this is once you have 20, 50, 100 or more variables to solve for. Are you currently in H.S. or college?



A few things that I have learned over the years that may help are as follows:



Clearly write out equations and define variables. So many people just scribble down some numbers on scratch paper and can't really see what they are doing.



If you can't figure out a problem (like during a test), write down as many of the equations, etc. that you can and explain how you think the problem may be solved. Teachers want to see that you understand the concept more than getting the actual correct answer.



Clearly show work for all problems, even if the teacher doesn't ask. It will help to make sure that you did the work correctly and the teacher will know that you did the work. It will also help get you partial credit if you made a mistake. For instance, if you got a sign wrong or multiplied instead of dividing during the problem and got the wrong answer. If you just wrote the answer it would be wrong and the teacher doesn't know where you made the mistake. If you write out your work, the teacher can see where you made the mistake and will often give partial credit, because they want to see understanding instead of just the right answer.



Keep track of units. This is important in physics and math problems that deal with different units (area, pressure, force, etc.).



Double check work. Once you solve the types of problem that you have been doing plug the answers into the original problem and see that it works (For instance the plane problem the plane is 150 mi/hr and the wind is 30 so 150+30=180 (going with the wind) 150-30=120 (going against the wind)) If the numbers don't match the problem statement, you probably made a mistake. Go back through the work and try to find the problem.



Does the answer make sense? This goes along with checking work. For instance, if you got the variables mixed up on the plane problem and found the plane speed to be 30 mi/hr and the wind speed to be 150 mi/hr, you know that doesn't make sense and there is a problem somewhere.



If you follow these guidelines, you will be much less likely to make simple mistakes and will likely do better in your classes. Hope this helps some. If you have any questions, just ask. Feel free to PM me anytime.



Kent

drftk1d

I might have hope for passing calc this semester!