how to find a mathematical equation?
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
974911
To answer this I must first explain what an equation is:\nAn equation is where 2 sides are equal to one another.  Ex. 4x + 13 = -91\n\nAn expression does not have the = sign.  Example: 4x + 13
952054
To find the equation, just substitute the 2 roots to the "Quadramagic formula" (...)\nx² - (p + q)x + pq = 0\nwhere p and q are the roots.\n\nHence, p = 7 + i and q = 7 - i.  We substitute:\nx² - [(7 + i) + (7 - i)]x + (7 + i)(7 - i) = 0\n\nWe add and multiply\nx² - (14)x + (7² - i²) = 0\n\nSince i² = -1, then\nx² - 14x + 49 - (-1) = 0\n\nthen,\nx² - 14x + 49 + 1 = 0\n\nTherefore, the quadratic equation is\nx² - 14x + 50 = 0\n\nTherefore, the correct answer is 1).\n\n^_^
710964
ab=6\n\nit is an equation because of =
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INSTEAD of giving you the answer directly -- here are the steps to solve and simplify.\n\nWRITE down the equations on separate lines -- leave about 4 lines between each problem.\n\nNOW -- draw connecting lines between the LIKE TERMS (for example -- in the first problem (4x+6-2x+6) the like terms are those that have the 'x' value with them and the other set of like terms are the constants with NO variables with them) -- I would use two different color pencils to do that)\n\nso add the 4x and 2x together (Answer is _x) -- then add the constants (those numbers without variables together (or +6+6)) together (Answer is __).\n\nNow put these two results together on the 3rd line.\n\n\nFinally, factor out any common terms in the results.\n\nTHAT is your final answer and the simplified expression.\n\n\n\n\nFOLLOW these steps once again for the 2nd problem -- but now instead of the variable being 'x' it is now 't'.\n\nNOW
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Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3
795656
The first thing you should remember is the phrase PEMDAS.  This is shorthand for parenthesis, exponents, multiplication/division, and addition/subtraction.  This is the order of operations in math.  You should solve all equations in this order.  So for the first one, you should first distribute (multiply) -7 with (3-3x).  After that combine like terms, i got 25x -30; you can also factor that to be 5(5x-6).  The same steps apply for the rest of the problems.  I hope it helps.
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to find x-intercept => you have substitute y=0 into the equation and solve for x.\n\nIn this case (0,0) is the x-intercept.
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T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
619519
The first one is a quadratic equation.\n\nLet y = 3 ^ x then \n\n==> y^2 -10y + 9 =0\n(y- 9)(y-1) = 0 \ny= 9 , 1  <--- you can get this from a variety of methods including the quadratic eq'n formula\n\n\nsince y = 3^x = 9,1\n then x  = 2 or 0  or  <-- can't explain a simple method to get that other than I just saw that.\n\n18 = 2^(x+3) + 2^x\n18= 2^x*2^3 + 2^x\n18= 8* 2^x + 2^x\n18= 9*2^x\n2= 2^x\nx=1
783117
Without getting into the graphing explanation, the simplest reason I have is that you can find X using the quadratic equation if the equation is in the form ax^2 + bx + c = 0. You set the equation equal to zero so it matches the form and then you can use the quadratic equation to find X.
923416
Sometimes I like math.  What you need to do is add 1 to both sides of the equation for x=3y-1\nyou will be left with (x+1)=3y you should then divide the 3 from both sides... afterwards you will have (x+1)divided by 3 equals y or (X+1)/3=y\nFor the other part x+2y=9 you should substitute (x+1)/3 for the "y"\nNow solve x + 2((x+1)/3)=9\nx+(2x/3) +(1/3)=9\ntake from there
55347
That's the answer, but that doesn't tell you how:\n\nwrite out the equation like such:\n\n(a+b)(a+b)\n\nNow use a method called the FOIL method.\n\nmulitply the FIRST part of each unit (F from foil)\n\na*a = a^2 (or a squared)\n\nNow multiply the OUTER part of each unit (O from foil)\n\na*b = ab\n\nNow muliply the INNER part of each unit (I from foil)\n\na*b= ab\n\nFinally mulitply the last part of each unit (L from foil)\n\nb*b = b^2 \n\nNow combine all the like terms: \n\nab = ab = 2ab\n\nLeaving  you with:\n\na^2 + 2ab + b^2   (or a squared plus 2ab plus b squared)
576570
algebra...usually (well i do) set up a equation for everything, and solve for the variable. If u need help u can always get help in www.algebra.com .....
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Next time just type your math equation into Google and it will do the math for you.\n\n15