How do you do these math problems???? I need HELP!! It's only a few problems!! PLEASE! I AM CONFUSED!!!?
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Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.
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Right lets see if this helps ......... \n a)\n4x + 4y = -12\n-2x + 3y = 14\n\nDouble the second equation and add:\n4x + 4y = -12\n-4x + 6y = 28\n10y = 16\ny = 16/10\ny = 8/5\n\nThen solve for x:\nx + (8/5) = -3\nx = -3 - 8/5\nx = -4 3/5\n\nb) \n3x + 2y = 5\n2x - 3y = 3\n\nMultiply the top by 2 and the bottom by 3, then subtract:\n6x + 4y = 10\n6x - 9y = 9\n13y = 1\ny = 1/13\n\nThen solve for x:\n3x + 2(1/13) = 5\n3x + 2/13 = 5\n3x = 63/13\nx = 21/13\nx = 1 8/13 \n\n2 points to me, that helps ha ah ha ha
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I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
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***************\nQUESTION 1\n***************\nOn the first question, we first have to find the equation that we will be solving for.  \n\nAssume that x is one integer.\nWe know that x+1 is the other integer because they are consecutive.\n\nSo we now put the wording into an equation\n1/3(x + (x+1)) = x - 5\n\nNow we solve the equation\n1/3(2x + 1) = x - 5\n2/3x + 1/3 = x - 5\n16/3 = 1/3x\n16 = x\n17 = x + 1\n\nOur two integers are 16 and 17.\n\nNow we check our answer.\n1/3 of 33 = 11 & 16 - 5 = 11 We are good! :)\n\n\n***************\nQUESTION 2\n***************\nEvery hour, Amy and Randy together walked 3.6 + 4.2 km apart from each other.\n\n3.6 + 4.2 = 7.8\n\nSo, every hour they will be 7.8 more km apart from each other.  We only want 10 minutes though, and there are 60 minutes in an hour. To get how far apart they will be in 10 minutes, we take:\n7.8 * 10/60 = 1.3\n\nRandy and Amy will be 1.3 km apart after 10 minutes.\n\nWe are done! :)
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(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
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Is it possible to have 2 equal signs in a problem?\nI am only in algebra but I haven't seen one like that before.\nSorry I couldn't help.
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This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
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The first problem is easy\nx=7/2 or x=3 and 1/2\n\nand for the second it is still pretty easy\nJust put in 0 for y or x which ever one you want to find\nLets find y first...so you put 0 in for x\n\n4(0)+5y=10\n5y=10     divide both sides by 5\ny=2\n\nThen find x....so put 0 in for y\n\n4x+5(0)=10\n4x=10       divide both sides by 4\nx=10/4      then simplfy\nx=5/2\n\nx=5/2    y=2\n\n\nohhhhhhhhhhh my bad I thought they were 2 different problems....well if they are 2 different problems im right....if they are 1 whole problem then the other people are right....
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Well. Do you know how to work rational expressions in general? The first link below is a useful general resource and has a section on rational expressions and polynomial long division; the second is specifically on rational expressions at college intro-level.\n\nI used to tutor math and I found that most people's difficulty with word problems was not the actual math, it was extracting the appropriate equations from the words. Once they had the equations, they could solve them (assuming they'd learned the basics). \n\nI'd find it pretty difficult to give you general guidelines on "how to solve word problems with rational expressions" but the main trick with word problems is to figure out a) what variables are involved, and assign each of them a letter (e.g., speed of first train is x, time it has been travelling is y, speed of second train is z...), b) sort out the relationship between them (e.g. x = 2z/5), and c) make sure you know what *exactly* is being asked. It is not at all uncommon for word problems to include extraneous information (colour of train, age of driver) to muddle the issue.\n\nYour best bet is almost certainly to go ask your teacher, professor, TA, tutor, study group, or classmate-who-understands-it for help. They will be better able to figure out where your problem is if they can watch you work through an actual example.
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All you have to do is combine like terms and then solve for x or y.  Then for the solving and graphing, refer to your textbook.  I don't feel like telling you HOW to graph it.
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U R RIGHTT..HOW DID YA GET IT..PLZ CAN YA EXPLAIN..I'VE BEEN TRYING THIS PROBLEM FROM A LONGTIME.EVEN DE CALCULATOR SAYS ITZ "4"..I KNOW DE CALCULATORS WRONG TOO..PLZZ DO EXPLAIN IT TO ME ASAP
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Go back to math class and figure it out!
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Option a: use one of the fancy calculators around nowadays.\nOption b:\nOk, we need at least a reasonable guess to get going. Lets draw a graph. But how do you draw a graph of 2cos(x) + x? We don't. We draw a graph of 2cos(x), and another graph of -x, and see where they are equal - because thats the point we're looking for. Don't forget to use radians, since you can't add a number to a degree.\nThe solution looks like it is around -1 or so. Now we can do a bit more guessing.\nPutting -1 into that equation gives 0.0806. Lets try -0.9. Whoops - that gives 0.34, which is getting even worse. How about -1.1? Ahha! That gives -0.19, so we know the answer is between -1.1 and -1.0.\nWe can keep narrowing it down in this way. Try -1.05: thats negative, so we're between -1.0 and -1.05. Try -1.02: thats positive, so we're between -1.02 and -1.05. Try -1.03: thats negative, but nearly 0. So we're between -1.02 and -1.03.\nWe're nearly there - try -1.025. thats positive, so we're done! We know the answer is between -1.025 and -1.03, so it must be -1.03 to the nearest 100th.\nThats quite a bit of work, but its not too bad. There are faster ways to do it, but they're a bit more complicated.
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Add them... 8+7+7+5+3+2+2= 34 and divided by 7\nYyou have 34/7=\nYou know 4x7= 28 so you need 6 more to reach 34\nso you have 4 and 6/7
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The best place to go for math help online that i found was \nhttp://www.math.com\nI hope its useful to you.  I know people on here can tell you the answers, but if you have a test to take, and want to really learn it, go there.
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i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
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Multiply or divide all components in one question to set the variables in the opposite questions to cancel out if you added the two problems together by combining like terms. For example:\n2x + 3y = -13  and 4x + y = -1\n\nTo keep this simple I will multiply all the components of the 2x + 3y = -13 equation by negative 2, resulting in:\n\n-4x - 6y = 26\n\nThen add the new equation to the other equation 4x + y = -1\n\n-4x -6y = 26\n4x + y = -1\n----------------\n0x -5y = 25  which simplifies to y = -5\n\nInsert the value found for y back into one of the original eqations and solve for x\n\n2x + 3(-5) = -13\n\n2x + -15 = -13\n\n2x = 2\n\nx = 1\n\nYour end result is x =1 and y = -5