i need help with x^2 + c^2 = 19c^2
124439
just like the person above (bree) says but you can further reduce the 18. \n\nx= 3c sqrt(2)
652672
Use pythagorean theorem\n(2x)^2 + x^2 = 10^2\n4x^2 + x^2 = 100\n5x^2 = 100\nx^2 = 20\nx = sqrt(20)\nx = 2sqrt(5)\nArea = (1/2)(2x)(x)\n= (1/2)(2*2sqrt(5))(2sqrt(5))\n= (1/2)(4*2*5)\n = 20
529747
(x+3)(x^2 + x + 2)\n= x(x^2 + x + 2) + 3(x^2 + x + 2)\n= x^3 + x^2 + 2x + 3x^2 + 3x + 6\n= x^3 + 4x^2 + 5x + 6
445320
sure. no problem, pal.\nx^2 + 9x + 20
192095
ok, u have to use distance formula\n\nsqrt[(-4-(-7))^2+(c-7)^2]=5   \nsquare both sides\n[(-4-(-7))^2+(c-7)^2]=25      \nadd the stuff in the parentesies\n[(-3)^2+(c-7)^2]=25           \nexponents\n9+(c-7)^2=25                  \nsubtract from both sides\n(c-7)^2=16                    \nsquare root both sides\nc-7= +/-4                     \nadd to both sides\nc=11 or c=3
519848
x^2 - 2(3-x^2)\nDistribute the -2\nx^2 + (-2)(3) + (-2)(-x^2)\nSimplify\n= x^2 - 6 + 2x^2\nCombine like terms\n= 3x^2 - 6
410796
put y = 7 and x =4 and solve for c\nwe get 7 = 16+c or c = -9
75947
[ 1x + 9 ] = 19\n[ 19 - 9 ] = 1x\n  10 = 1x\n  x = 10\nYou need to study!!!\nGood Luck in Math Class!!
347670
x(rt2)x+5rt2(x^2)-25y(rty)\n=6rt2x^2-25yrty
349671
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
471703
Here's what you have to do.\nReplace Q with 1.6x10^(-19)\nReplace C with 3.31x10^20\nReplace K with .49\nFor #1, replace x with 1.9\nF = (1.6x10^(-19)) (3.31x10^20) (.49) / (2(1+cosh(.49*1.9))\nEvaluate using a calculator.\n\nFor #2, replace F with 3 and solve for x.\n3 = (1.6x10^(-19)) (3.31x10^20) (.49) / (2(1+cosh(.49*x))\n3(2(1+cosh(.49*x)) = F = (1.6x10^(-19)) (3.31x10^20) (.49) \n6 + 6cosh(.49*x) = F = (1.6x10^(-19)) (3.31x10^20) (.49)\n6cosh(.49*x) = F = (1.6x10^(-19)) (3.31x10^20) (.49) - 6\ncosh(.49*x) = ((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6\n.49*x = inverse cosh (((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6)\nx = inverse cosh (((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6) / .49
651316
Set 16t^2 +32t -225=0\n\nUse quadratic formula gives t = 2.9sec
564274
[sqrt(2)]^2 = c^2 - b^2\nLet c = 3/2 and b = 1/2, then\n[sqrt(2)]^2 = (3/2)^2 - (1/2)^2\n2 = 9/4 - 1/4\n2 = 8/4\n2 = 2
65980
Start with adding 7 + (9 + 12).  The sum is 28.  The equation now looks like this:\n\n(c + 9) + 12 = 28\n\nNext, subtract 12 from both sides of the equation:\n\n(c + 9) + 12 - 12 = 28 - 12\n\nThe equation now looks like this:\n\nc + 9 = 16\n\nSolve it from there.
139665
3(x^4+3x^2-11)\n\nseriously you need to pay attention in class instead of asking people. lol
183477
Use remainder theorem and polynomial division for the first one and then either the quadratic formula or, if you know your maths, logics. Just plot the second one in a graph and you'll see the answer. The next few lines I just don't understand what you are writing, perhaps if you wrote it more clearly. As for the simplify bits, you don't really need help there, it's a matter of seeing how you can create similar terms. The last bit, once again, I do not know what you mean. \n\nEdit: corrected some spelling mistakes (it's quite late over here).
404469
7x^2+16x over (x-2)(x+2)(x+3)