
#1
Jan1005, 11:17 AM

P: 2

Hi
i am in us military and over the years i have lost my mathematical skills so picked up a book from library to solve simple equations.. i couldn't figure out few problems.. i will show you my work and then you guys can explain where i messed up. i have the correct answers but that does not matter i would like to understand the concept. Some of these problems are elementary level but i have not done this for a while. There are no Math teachers here i can get help from so you guys are my only help, i am currently stationed in Turkey and Probably going to be in Iraq a lot from here on out. I also try not to use the calculator because i really suck at math so harder it is better i will get at it so please just correct my simple mistake although I did check my work over but i could be wrong about most simple things. thank you very much for your help here are the problems 1. A = 1/2h(B1 + BA2) Solving for H A= 1/2ba1 + 1/2ba2 1/2ba1A = 1/2ba2  1/2BA1A 1/2BA1A H =  hB2 / B1A thats what i got but its wrong correct answer in the book is = 2a / B1 + B2 I tired it several times please explain step by step how to arrive at the correct answer. 2. 1x + 2(1/6x + 2) = 6/5x + 16 Solve for X 90 1x + 1/3x + 4 = 6/5x + 16 4/3x = 6/5x + 12 < i think i messed up here somewhere in the fractions i am not good with those. 4/3x  6/5x = 12 some How X = 90 in the book i got stuck at this problem so i need help on this one too.. i suck at fractions i am reviewing that right now. I didn't know how to make that PIE symbol on this so i wrote "#" Also on this one i know its a minor mistake because i probably forgot how to divide or multiply a fraction or exponintial value. # = pie 3. v= 1/3#*h^2(3Rh) Solve for R V= 3R*1/3#*h^2  1/3#*h^3 V + 1/3#*h^3 = 3R*1/3#*h^2   1/3#*H^2 1/3#*h^2 V + h^1 = 3R   3 3 R = v + h^1  3 Correct answer in the book is R= v + 1/3#*h^3 3v + #*h^3  =  #*h^2 3*#*h^2 you can explain much as you like more the better thank you Max 



#2
Jan1005, 12:17 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

"1. A = 1/2h(B1 + BA2)
Solving for H A= 1/2ba1 + 1/2ba2 " Was there an A1 in the original equation? If not where did a1 come from in the second? Or are BA1 and BA2 indivual symbols? What became of the "h" that you were supposed to be solving for? And, by the way, it is a very bad idea to use A and a to mean the same thing. In some formulas they are used to mean different things and it's confusing! Also does 1/2h(BA1+ BA2) mean (1/2) h(BA1+ BA2) or (1/(2h))(BA1+ BA2) or 1/(2h(BA1+BA2)) ? Since you are restricted to one line here, please use parentheses to clarify. If the equation is A= (1/2)h(BA1+ BA2) then just divide both sides of the equation by (1/2)(BA1+ BA2) to get h= (2A)/(BA1+ BA2). If the equation is A= (1/(2h)(BA1+ BA2) then first multiply both sides by 2h to get 2hA= BA1+ BA2 and then divide both sides by 2A to get h= (BA1+ BA2)/(2A). If the equation is A= 1/(2h(BA1+BA2)) then multiply both sides by h to get Ah= 1/(2(BA1+ BA2)) and then divide both sides by A to get h= 1/(2A(BA1+BA2)). In order to get the answer you say the book has, the original equation must have been simply A= (1/(2h)(B1+B2) which is similar to my second suggestion but where did you get the "a" or "A" in the right hand side? If A= (1/(2h))(B1+ B2), as above, multiply both sides by 2h to get 2hA= B1+ B2. Now divide both sides by 2A: h= (B1+ B2)/(2A). Not to put too fine a point on it, to do mathematics you must be carful and precise. If you are not even able to copy the problem correctly, you are not giving it enough attention! "2. 1x + 2(1/6x + 2) = 6/5x + 16 Solve for X 90" X 90 ??? Another typo? Do you mean solve for x? (Oh, I see, you had copied the answer.) I'm going to assume that 1/6x mean (1/6)x and 6/5x means (6/5)x. Multiply out 2((1/6)x+ 2)= (2/6)x+ 4= (1/3)x+ 4. Now you have x+ (1/3)x+ 4= (6/5)x+ 16. Subtract (6/5)x and 4 from both sides. That will get everything involving "x" on the left and everthing that does not have an "x" on the right. You will have x+ (1/3)x (6/5)x= 16 4 which is the same as (1+ 1/3 6/5)x= 12. This is, in fact, exactly what you have with "4/3x  6/5x = 12". Now there are two ways of proceeding: this is (4/3 6/5)x= 12 and a "common denominator for 3 and 5 is 3*5= 15. 4/3= 20/15 and 6/5= 18/15. 4/2 6/5= 20/1518/15= 2/15. (2/15)x= 12 so divide both sides by 2/15. That's the same as multiplying by 15/2: x= 12(15/2)= 6(15)= 90. If you really don't like fractions, the other way of proceeding is to multiply both sides of (4/3)x  (6/5)x = 12 by both 3 and 5 to cancel the denominators: (3)(5)(4/3)x (3)(5)(6/5)x= 20x 18x= (3)(5)12= 180. 20x 18x= 2x= 180 so x= 90. 3. v= 1/3[itex]\pi[/itex]h^{2}(3Rh), solve for R Again, I'm going to assume that this is v= (1/3)[itex]\pi[/itex]h^{2}(3Rh) To solve for R, first divide by (1/3)[itex]\pi[/itex]h^{2} to get that "away" from the R in (3Rh). Now we have (3v)/([itex]\pi[/itex]h^{2})= 3R h. Add h to both sides and we have 3R= (3v)/([itex]\pi[/itex]h^{2})+ h We can combine those by getting the same denominator: [itex]\pi[/itex]h^{2} and that means we will need to multiply the "h" by that: h([itex]\pi[/itex]h^{2})= [itex]\pi[/itex]h^{3}. 3R= (3v+ [itex]\pi[/itex]h^{3})/([itex]\pi[/itex]h^{2}) so finally, R= (3v+ [itex]\pi[/itex]h^{3})/(3[itex]\pi[/itex]h^{2}) 


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