how do you do a pre alegbra equation on general first degree what are the steps?
353406
You will want to isolate the variable (many times it is 'x'). First you will add or subtract from each side. Then divide. For example:\n5x+4=19, subtract 4 from each side giving 5x=15. Then divide by 5 giving x=3. I hope this helps.
704007
use arithmatic progression formula n(2a+(n-1)d)/2\nwhere n=no. of terms. a=value of first term and d=common difference between each consecutive term\nsay fro your first eg. no. of black ball = 1+3+5+...\nn=10 a=1 d=2\nso sum = 10(2+9*2)/2 = 100\nsolve the rest by changing values of n, a and other variables respectively
878075
First of all let's see the expression of Kp for the first equation\n\nH2+Br2 <=> 2HBr\n\nKp=P(HBr)^2 / (P(H2)*P(Br2))= 3.5*10^4\n\na) HBr <=>1/2H2+ 1/2Br2\n\nKp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)\nBut if you look carefully  Kp1=1/squareroot(Kp) =\n=1/(SQRT(3.5*10^4))= 5.35*10^-3\n\nRemember that since squareroot of x= x^1/2\nKp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)=\n\n= [P(H2)*P(Br2)/(P(HBr)^2)] ^(1/2)=\n\n=1 / [P(HBr)^2/(P(H2)*P(Br2))] ^(1/2) =\n\n=(1/Kp)^(1/2) = 1/Kp^(1/2)=1/Squareroot(Kp)\n\nI am sorry I couldn't write it more clearly but it is a limitation of the site and we can't write equations in a better way.\n\nb)2HBr <=> H2+ Br2\nIt is reverse of the initial reaction so\n\nKp2=(P(H2)*P(Br2)) / P(HBr)^2= 1/Kp= 1/(3.5*10^4)= 2.86*10^-5\n\nc)1/2H2+1/2Br2 <=>HBr\n\nis the reverse of a do Kp3=1/Kp1= 1/(5.35*10^-3)=1.87*10^2
343985
need to use the chain rule i.e.\n\nf'(pq) = p'*q + p*q'\n\nwe get:\nae^ax/(sin(bx+c)) - be^ax*cos(bx+c)/[(sin(bx+c))^2]\n\nThis is the correct answer\n\n*edit* srry couldnt see all of it if its on one line so split over two lines the answer is:\n\nae^ax/(sin(bx+c)) \n-be^ax*cos(bx+c) / [(sin(bx+c))^2]\n\n\nand ofcourse if you like, further simplification is possible with cos(bx+c)/sin(bx+c) =1/tan(bx+c)=cot(bx+c), also you can take a factor of e^ax out if you like but the above answer is perfectly accebtable.
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
723293
you have to use the couchy schuartz method\nim nit sure on the spelling \nbasicly you take one vector (V1) and use it then you take the second vecter (U1) minous vi times(V1 interproduct U1)/(magnatude V1^2) and use that total as V2 then you do a simular thing for the third and fourth vector and up if you have them by subtrating V1<Ua,V1>/(mag V1)^2 plus V2<Ua,V2>/(mag V2)^2 .... till you have done that for each vector \nin other words every vector will have a term for all of the new previous vectors subtracted from it then finaly you take all the new vectors and devide them by their magnatude to get unit vectors and they will then all be orthOGANOL UNIT VECTORS
1079811
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
822981
i)Just substitute in the values to get F\nx = 1.1.  Other values are given.  Hence F.\n\nii) Rearrrange the equation\nx = [cosh^-1(QCK/2F-1)]/k\nF = 2.9.  Other values are given, hence x.
616179
here are the general steps you want to take when approaching these math problems\n\n1.) check for the greatest common factor\n2.) write out your parenthesis\n3.) since it's a squared problem, fill in the x's\n4.) write out your + & - signs\n5.) list the multiplication factors for the last term\n6.) plug in factors that will lead to the correct product\n\nwe'll start with #1 to get a better idea\n\n1.) -x^2+5x+24\n\n1. there is no greatest common factor\n2. write out your parenthesis\n\n(    )(    )\n3. fill in the x's\n\n( x   )(x   )\n\n4. fill in + or - signs\n\n-since we're dealing with -x^2 we know that one of the x's need to be + and the other one -\n\n(x   )(-x   )\n\n-since our last term is positive we know that either both the #'s are + or both of them are -\n\n(x-  )(-x-  )\n\nor\n\n(x+  )(-x+  )\n\n\n\n5.  list the multiplication factors for the last term\n\n24\n\n1*24=24\n2*12=24\n3*8=24\n4*6=24\n\n6. plug in factors that will lead to the correct product\n\n(x-8)(-x-3)=-x^2+5x+24\n\n**now, with a little more practice, you'll be able to figure these out faster...here are the answers to the rest of the problems\n\n2. (x-12)(x-12)\n3. (x-9)(x-12)\n4. 2(x^2-8x+16)=2(x-4)(x-4)\n5. 3(x^2-6x+72)=3(x-12)(x+6)
282632
simple, apply the cubic formula.
300751
Use the chain rule and the power rule.\n\nIf y = x^a, then y' = a * x^(a-1).\n\nSo, y = (2x+3)^(.5)\ny' = .5 * 2 * (2x+3)^(-.5) = (2x+3)^(-.5)\ny'' = -.5 * 2 * (2x+3)^-1.5 = -(2x+3)^(-1.5)
234536
Okay, equations of the form ax^2+bx+c=0 have the discriminant b^2-4ac. In your case a=4,b=-3,c=5. So you have to take (-3)^2-4*4*5=-71. Since the discriminant is negative, you have 0 real solutions. If it were 0, you would have 1 real solution, and if it were positive, you would have 2 real solutions. Hope that helps!
678914
Use FOIL\nF-first terms: x(x) = x^2\nO-outer terms: x(-4) = -4x\nI-inner terms: 3(x) = 3x\nL-last terms: 3(-4) = -12\nAdd everything together and combine like terms to get\nx^2 - x - 12
538371
http://en.wikipedia.org/wiki/Cubic_equation
831482
We just started this stuff in school.  first, think of f(x) as y.  then make a table with the left column as x, and the right column as y.  the vertex is going to be -b/2a.  so in this case.  it'll be -1/4.  that's the middle number in the table and also the vertex.  now put above that number in the table on the x column as so.\nx  y\n1\n0\n-1/4\n-2.5\n-3.5\nnow fill in the the x column by plugging in the x variables.  plug in 0 for x and thats your y-intercept.  then plug in your 0 for y, and thats your x-intercept.  this is going to form a parabola so the axis of symmetry is -1/4.  the domain is going to be all real numbers because there are no contradictory numbers to fill in.  Im not sure how to do the range because im only a junior in high school but I hope I helped a little.
208740
EDIT: ** Baah. When I typed this answer I was first. Then at the end, I'm like the 3rd. Oh well..**\n\nThere are two ways. Which one is easier depends greatly on your thinking skills.\n\nYou can take the expression x2+7x+12 and assume it should break up in two factors:\n(x+ ? )(x+ ?)\nThen, search for two numbers that add up to the 7, but when multiplied equal 12.\nIt helps if you break down 12 into it's components, 12= 2*2*3. So, obviously, the answers could be:\n\n2 * (2*3) = 2 * 6 (but 2+6 is not =7)\n3 * (2*2) = 3 * 4, voila! 3+4 also = 7, so:\n\n(x+3)(x+4).\n\nThe second way could be just solving the cuadratic formula for second degree equations:\n\nb=(-b +/- SQRT(b2 - 4ac))/2a\n\nWhere ax2 + bx + c, so you'll have: a=1, b=7, c=12.\n\nReplace, then solve twice: Once for the + value of the SQRT, another for the - value of the SQRT. You should get something like:\n\nFirst run:\nx= -3\n\nSecond run:\nx= -4\n\nx=-3 is the same thing as (x+3)=0\nand \nx=-4 is also (x+4)=0\n\nNotice the two factors are precisely (x+3) and (x+4)?\n\nHope this helped..