how do you use the method SOHCAHTOA?
544028
You have to have a right triangle.\nThen you figure out which is side is opposite or adjacent from the angle you need.
985667
They taught us SOHCAHTOA back in OL trig, but I never really felt the need for it, I mean just picture a right angled triangle and the longest side is the hypotenuse,h. Take an angle and the side opposite it is h.sin x , where x is the angle. The other side is h.cos x
1122155
not sure what box method is...the answer is\n(2x+5)*(x-3)
563324
There are a couple of methods but the one I like to use is the method of half reactions.  First we split the reaction into an oxidation half reaction and a reduction half reaction as follows:\n\nCr2O7(-2) -> Cr(+3)   and    CH3OH -> HCO2H\n\nWe know that the left half reaction is the reduction half because the oxidation number of the Cr changes from (+6) to (+3).  Likewise the right reaction is the oxidation because the oxidation number of the carbon changes from (-2) to (+2).\n\nNow that we have our half reactions the next step is to balance everything EXCEPT oxygen and hydrogen.  In this case the only other elements involved besides O and H are Cr and C.  Doing this give us: (the right one is already balanced for carbon; one on each side).\n\nCr2O7(-2) -> 2Cr(+3)   and    CH3OH -> HCO2H\n\nStep three: We balance oxygen be adding water molecules to the side with less oxygen atoms.  Doing this give us:\n\nCr2O7(-2) -> 2Cr(+3) + 7H2O   and   H2O + CH3OH -> HCO2H\n\nStep four: Since we are in acid we add H+ ions to balance the hydrogen atoms on both sides.\n\n14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O      and         \n\nH2O + CH3OH -> HCO2H + 4H(+1)\n\nStep five: We balance the charges on both sides by adding electrons [e(-1)] to the side with the most positive charge.  Looking at the first half reaction we see that there is a total charge of +14-2 = +12 on the left side, and a total charge of 2*(+3) = +6 on the right side.  Therefore adding 6e(-1) to the left side should balance out the charges giving us: (do the same for the second half reaction)\n\n6e(-1) + 14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O    and\n\nH2O + CH3OH -> HCO2H + 4H(+1) + 4e(-1)\n\nYou know you have made a mistake if the electrons don't end up on different sides of the two half reactions.\n\nStep six: We must balance the two half reactions to each other i.e. the number of electrons produced in one reaction must equal the number of electrons consumed in the other reaction.  If we multiply everything in the first reaction by 2, and everything in the second reaction by 3 this will give us 12 electrons in both half reactions.  Doing this gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) -> 4Cr(+3) + 14H2O    and\n\n3H2O + 3CH3OH -> 3HCO2H + 12H(+1) + 12e(-1)\n\nStep seven: Add the two half reactions together and simplify the result as much as possible by cancelling things that appear on both sides, and by finding the smallest set of whole number coefficients for the reactants and products.  Adding the half reactions together gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1) + 12e(-1)   \n\nCanceling the electrons on both sides gives:\n\n28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1)\n\nCanceling the extra H+ ions on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H\n\nCanceling the extra H2O molecules on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3CH3OH -> 4Cr(+3) + 11H2O + 3HCO2H\n\nIt looks like this is the most simplified version.\nLast step: Check to see that the final reaction is balanced for both atoms and charges.  It looks like this is the final answer.
991404
Here is one way:  http://www.ulc.net/index.php?page=ordain&PHPSESSID=c47850e11956c4bc515af318d5bb5728\n\nThis is how I did it.
255619
http://www.anywho.com
1082211
was difficault method, choose Hermit is better
41764
you can find a very good tutorial at the link.\n\nNo need to write whats already writtenm =)
961641
http://hamilton.bell.ac.uk/hncexam/Samplepaper.doc ;-)
589334
http://aleph0.clarku.edu/~djoyce/mathhist/mathhist.html
186264
http://www.howtomakeamilliondollar.com
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
886349
SOHCAHTOA \nSOH =sine=Opposite /Hypotenuse\nCAH cos=adjacent/hypotenuse\nTOA tan=opposite/adjacent\nIn finding the adjacent,we use CAH\ncos19=adjacent/13.8\nadjacent=cos19x13.8\nadjacent=13.048
376678
try the Bates method and pinhole glasses
503650
Why don't you share with us what you have done so far instead of just asking someone to do your homework?\n\nSorry about the snide remark. Here is your well-deserved step-by-step solution:\nhttp://www.rgetso.com/yahoo/afchica101.pdf
76773
http://www.rom.on.ca/egypt/hiero/hiero.html\n\nGive it a try!  Have fun.
838334
=I found a website for you..  If you click on the link below they will send you a free brochure which explains the techniques used.