What could i write for an equation for this translation: ( left 2 units ) when y = I x I ?
152046
y=|x+2| \n this shifts the "v" shaped graph 2 units to the left from the origin.
408980
when playing with < and >\nif you multiply all the equation by -1 the sign < changes to >, and the sign > changes to <\nbut if you just switch the number on the other side the sign doesnt change\nin ur case :\n2x-y < 4 \nequals\n-y < -2x + 4\nif you multiply it with -1 it will become\ny > 2x - 4\n\nlike that. hope it helps :)
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
179147
ur ques should be\ny = sq root x + 8 ryt?\nknow i leave it to u to find the grapg of the sq. root func ok\ntalking about the translation, c lets take y = sq r x\nnow when x = 0 then y is also = 0\nbut now if we add 8 to it ie y = sq r x +8, then at x = o y=8\nthats the trtanslation\nthat means the whole graph shifts by 8 unit on the +y axis
836116
Let x and y be your numbers.  Your equations, therefore, are the following:\n\nx+y = 72\nx-y = 12\n\nDo you need to actually solve for x and y as well, or just set up the equations?  If you also need to solve, then write your two equations one on top of the other, then subtract to get an equation in one term only:\n\nx + y = 72\nx - y = 12\n----------------\n0x + 2y = 60\n\nSo, 2y = 60, which means y = 30\n\nNow substitute y=30 into one of the original equations to solve for x:\n\nx + y = 72\nx + 30 = 72\nx = 72  30 = 42\n\nSo your two numbers are 30 and 42\n\n~ ♥ ~
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
944317
Im guessing your solving for y... so make the problem 58=29Y.  Divide both sides by 29 and you get y=2
679588
Multiply it out...\n2y+6y+24x-24y-6x\nThen combine\n-16y+18x OR\n18x-16y
1147102
pi*(r^2)\n\nIt doesn't matter which unit you use, just make sure you're consistent.
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
412953
An equation that is incorrect.
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
974937
please complete the equation
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x