6 problems please help.?
964688
1/2(3x+2)=13\n(3x+2) = 26\n3x= 24\nx= 8\n\n-2(6-5a)= -8 \n6-5a= -6\n-5a=-12\na= 12/5\n\n-8(3p-2)=8 \n3p-2=-1\n3p=1\np=1/3\n\n5(7+n)=36\n7+n=36/5\nn= 36/5 - 7\nn= 1/5\n\n8y-(3-6y)= -1 2/3\n8y+3+6y= -1 2/3\n14y= -5/3 - 9/3\n14 y = -14/3\ny= -1/3\n\n2(3b+4)-(b-5)=12\n6b+8-b+5=12\n5b=12-13\nb=-1/5
794018
you should read your textbook and do practice problems to help.  ask your teacher or classmate for help.
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
920147
do your own homework. at least try to make the question not sound like a homework question
232213
I'll do the fourth problem.\n\n-3x< -6  or x+5< -2\n\nfirst deal with one half- \n-3x< -6 (now divide by -3)\nx>2 (when you divide by a negative the sign is reversed)\n\nthen the second half- \nx+5<-2 (get x alone, so subtract five from each side)\nx< -7\n\nso you have:\n2 < x  or  x < -7\n\nto graph this there is a line with an arrow to the left of the first slash and another line to the right of the second slash\n\n<-----/-------0--/-->\n____-7______2
443281
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
1076304
(1) 1.3\n(4) 4\n(5) Right\n(6) 0\n(7) 0\n(8)0\n(9)12
573142
Type the problem on http://www.google.com and then click the link on top called GROUPS. You can find a lot of useful help in there.\n\nAlso try http://www.vbforums.com/
509467
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
1184666
(1) C\n   B = D + 2\n   M = B + 4\n   M = ( D + 2 ) + 4\n   M = D + 6\n\n(2) E\n   B = 2H\n   M = B + 6\n   M = ( 2H ) + 6\n   M = 2H + 6\n\n(3) D\n   12 bulbs / x minutes\n   12 bulbs / ( x minutes ( 1 hours / 60 minutes ) )\n   12 (60) bulbs / x ( 1 hour )\n   ( 2 (12) (60) ) bulbs / x ( 2 hours )\n\n(4) D\n   Stop 1: x\n   Stop 2: x + y\n   Stop 3: ( x + y ) / 3\n   Stop 4: ( ( x + y ) / 3 ) + 2 = ( x + y + 6 ) / 3
812688
1*6N right\n2*No Direction\n3*How much force and where\n4*???\n5*???
655880
for the first problem . . .\n\nadd 3 to each side\n    6^(-.2x)=10\n\nrewrite equation in log form\n    log(base 6)10 = -.2x\n\nuse the change of base property on the left side\n   (log10)/(log6) = -.2x\n\ndivide by -.2\n    x = (log10)/[(log6)(-.2)]\n    x = -6.425\n\n\n\nand for the next one . . .\n\ndivide by 7\n    log(base 5)x-3 = 15/7\n\nrewrite in exponent form\n    5^(15/7) = x-3\n\nadd 3\n    x = 5^(15/7) + 3\n    x = 34.462\n\n\nhope that helps!
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]