i need help with pre-college algebra?
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Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
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Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
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Well, while the first answer is right, that probably doesn't answer your question if you're struggling.  \n\nI could try to explain, but your best bet would be face-to-face, one-on-one tutoring.  \n\nMost teachers (I stress MOST) are really understanding and really WANT you to learn.  They will usually stay after or come before school to meet with you to help.  If your math teacher won't do this, ask your counselor (not a math teacher, I know, but surely has been to college and had to take math courses). Just tell them your situation, and stress to them that you really WANT to learn it.\n\nOnce you get the basic understanding, they're not that bad.  \n\nGood luck, sugar.\n\n\n********\nPS  Jedi's answer is along the right track, but 2/3 is NOT the same as 2/9...\n\nYou have to take your denominator (the bottom number) and figure how many times it goes into 9.  That would be three times.  NOW you have to take that 3 and multiply it by the numerator also (the top number)... which would give you six (2 x 3)\n\nLike this (keep in mind that 3/3 is the same as 1, so when you multiply 3/3 by something, you're not changing the value of the first number, only the way it looks)...\n\n2/3 x 3/3\nmultiply across;  2 x 3 = 6\n                                 3 x 3 = 9\n\nnow you have 6/9, and since the bottom number is the same now as your 8/9 in the problem, you can subtract it from the 8/9.\n\n8/9 - 6/9 = 2/9\n\nGet it?
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Remember this phrase and you'll be fine!\nPlease = Parenthesis\nExcuse = Exponents\nMy = Multiplication\nDear = Division\nAunt = Addition\nSally = Subtraction\n\nNow let's work on the problem.\n\nWe can see that Parenthesis come first, however, in this problem, they terms in parenthesis are unlike terms, therefore, we move on.\n\nThere are no exponents, so we move on again!\n\nMultiplication.  Ignore the 28 for a moment and concentrate on \n-6(2x+4).\n-6 times 2x is -12x and -6 times 4 is -24. Now we have this problem:\n\n28-12x-24=124\n\nLet's combine like terms.\n\n28-24= 4\n\nNow we have 4-12x=124.\n\nsubtract 4 from each side.\n\n-12x=120\n\nnow divide each side by -12\n\n-12x/-12= x and 120/-12= -10\n\nso x= -10\n\nHope that helped!!
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No problem, i just finished a calc3 course in college.\n\nYour problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)\nYou have to use the chain rule so you get\n125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)\n\nthen you can simplify how you like from there
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You have to do the equations in the [ ] first.\nThen you have to do multiplication before addition.\nHere's the order of how you should do math like this:\nP - Parenthesis\nE - Exponents (a number to the something power)\nM - Multiplication\nD - Division\nA - Addition\nS - Subtraction\n\nSo you do [2+6(-3)] first\n[2-18]\n[-16]\nNow it's 15-3[-16]\n15+48\nAnswer: 63
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If you can't get an answer here, try asking your Algebra teacher for extra help after school tomorrow.  It's sure to add to your grade that you asked for the help.
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Wow, there is a lot here!  I'll start with some tips for solving algebra problems.  The first thing you should always remember is to play the equal game.  Whenever you add, subtract, multiply, or divide, you should do the same to the other side.  The other tip is to use the order of operations when simplifying.  Remember the sentence: Please excuse my dear Aunt Sally.  The initial letters of each word stands for the order of operations.  Do the Parenthesis first, followed by the Exponents, then Multiplication or Division, and Addition and/or subtraction next.  Finally, you should always simplify first using the order of operations before you start shifting numbers from one side of the = to the other.  \n\nI'm not going to work all of these problems, but I'll show you a few to get you started.\n\n1)\n3x-7 = 2   Add 7 to both sides\n3x = 9   Divide by 3\nx = 3\n\n4)\n6x - 3 = 2x - (4-3)/3   Simplify the stuff inside the parenthesis first\n6x - 3 = 2x - 1/3  Add 3 to both sides\n6x = 2x - 5/3  Subtract 2x from both sides\n4x = -5/3  Multiply both sides by 3\n12x = -5  Divide both sides by 12\nx = -12/5\n\nI hope this helps.  Good luck!
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alright, you need to understand this much more than you need the correct answer. You won't have Yahoo! answers to help you on any tests.\n\nYou need help factoring. A real-life tutor who can talk to you and show you in person is best, but I'm going to try to help you understand. \n\na polynomial is in the form ax^2 + bx + c. This is from the formula of (Z*x + q) * (R*x + p). Don't let all these letters scare you. They are just place holders for numbers. \n\nThe only thing that can get the letter A in Ax^2 +bx +c is the first two parts of the "terms" you need to multiply together. With letters a= z*r. In order to get C, you only need to worry about the last parts in the terms. Using letters c= q*p. B is what makes sure you know what two terms were multiplied together. B = z*p + q*r. In order to "factor" out a term from a polynomial you have to find the set of z, p, q, and r that works. I'll show you now with numbers.\n\nLet's say you have the equation 2x^2 + 8x +8. See, it's in the form of ax^2 +bx + c. a=2, b=8, c=8. NOw you have to figure out the facotrs of 2 and 8 (which are a and c)\nfor a=2 the numbers that are multiplied together to get two(the factors) are 2 and 1. Since you know that only the first two parts of the things you multiply together have to get you a, you can set up something that looks like this: \n(2x + ?) *(x + ?) \n\nNow you need to find what two numbers fit into your equation to get you c=8. The factors of 8 are 1*8 or 2*4. Let's try putting 1*8 into the equation.\n(2x+1)* (x+8) = 2x^2 +17x + 8. That's not right, so lets try to reverse them.\n(2x+8)* (x+1) = 2x^2 +10x + 8. That's not right either, so we know only the factors of 2*4 will work in this equation.\nLet's try them\n\n(2x+2)*(x+4) = 2x^2 +10x +8. It may seem like we've tried each of the factors, but we can still reverse 2*4.\n\n(2x+4) * (x+2) = 2x^2 +8x+8. That's right. Once you get more practice and understand this better, you can "see" certain factors that have to be included just because "factoring" will be automatic. But, just to make sure you should always check you answer like I did when I showed you the example.\n\nNow that you hopefully understand the hard part, perfect squares are easy. All you do is take  the square root of 'a' and take the square root of 'c' and put them into the equation. When it is a PERFECT SQUARE this works because of: ax^2 + bx +c = (z*x+ q)^2. ax^2 = (zx)*(zx) so a = z^2, and the square root of a = z. c = (q)*(q) so c = q^2, and the square root of c= q. I'll use numbers now.\n\n100x^2 + 240x + 144. a = 100, so the quare root of 100 = 10. z=10. Then the square root of 144 is 12, so q=12. That means y=(10x+12)^2. Now, to find b, all you have to do is foil it out.
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1.  add 1 toboth sides, then multiply by 3\n\n(x+4)^2=18  now take the square root of both sides\n\nx+4= +/- 3sqrt2\nx=-4 +/- 3sqrt 2\n\nyou also could use the quadratic formula\n\n\n2.  multiply num and denom by 3 - 2 sqrt3\n\n12-8sqrt3 +2sqrt3 -2sqrt6  all over -3\n\n\n3.  mult num and denom by 2 - sqrt 12 and get \n\n12-6sqrt12 all over -8\n\n12-12sqrt2  all over -8\n\n3-3sqrt2 all over -2
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I can help, I've taken College Algebra, Calc. I, Calc. II and I've also taken many other math courses, and I'm also studying to be a computer programmer, so if you need help you can contact me at assassin7262002@yahoo.com
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http://www.math.com/homeworkhelp/PreAlgebra.html\nhttp://www.gomath.com/algebra.html\nhttp://www.softmath.com/algebra_stats4/Prentice-Hall-Mathematics-Pre-Algebra-practice-work-book.html\nwww.amazon.com\nwww.ebay.com\nFind books on amazon and ebay new or used for Pearson Prentice Hall Mathmatics\nGood luck.
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Just do a search for "algebra." You'll find a lot of web sites, including:\nhttp://mathworld.wolfram.com/topics/Algebra.html\nhttp://www.purplemath.com/\nhttp://www.algebrahelp.com/
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Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
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Try going to this website. The will explain it step by step and you can ask all the math questions you want. www.algebra.com
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http://www.math.com/homeworkhelp/Algebra.html\n\nhttp://www.mathsisfun.com/number_algebra.html\n\nhttp://mathforum.org/library/drmath/sets/high_alg_toc.html\n\nhttp://www.tellmehowto.net/answers/maths.html\n\nTheres a couple, First two are aimed more towards algebra; Hope you find what you need!
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http://www.sparknotes.com/testprep/books/newsat/chapter19.rhtml\nhttp://www.algebrahelp.com/\nhttp://www.gomath.com/algebra.html#algebra\nhttp://ca.search.yahoo.com/search?p=algebra+%2B+practice&ei=UTF-8&fr=FP-tab-web-t340&x=wrt\nhttp://www.glencoe.com/sec/math/studytools/prealglblr.shtml\nhttp://library.thinkquest.org/20991/quizzes/index.html\nhttp://ca.search.yahoo.com/search?p=algebra+%2B+quizzes&ei=UTF-8&fr=FP-tab-web-t340&x=wrt&meta=vc%3D\n\ngood luck!