2 easy Math Questions!?
963839
1) two hours from 9 to 11. so distance 45*2 = 90 km\n\n2) The car has done 100+108= 208 km in 5 hours\nthe speed is 208/5 = 41.6 kph
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
809433
1.rectangle\n\n2.trapezoid
824758
Question 1 : 4^x - 2 * 5^(2x) < 10^x\n\nFactorise :\n2^(2x) - 2 * 5^(2x) < 2^x * 5^x\n\nDivide through by 2^(2x) :\n1 - 2 * (5/2)^(2x) < (5/2)^x\n\nLet y = (5/2)^x\n\nTherefore, 1 - 2y^2 < y\nor, 2y^2 + y - 1 > 0\nor, (y + 1)(2y - 1) > 0\nThus, y < -1 or y > 1/2\n\nThat is, (5/2)^x < - 1 or (5/2)^x > 1/2\n\nBut (5/2)^x is always > 0, so the only\nsolution occurs when (5/2)^x > 1/2.\n\nTake the reciprocal of both sides :\n(2/5)^x < 2\n\nTake logarithm of both sides :\nx(log2 - log5) < log2\n\nTherefore, x < log2 / (log2 - log5)\nor x < -0.7565 approximately.\n----------------------------------------------------------------------\nQuestion 2 : 7 * 3^(x + 1) - 5^(x + 2) = 3^(x + 4) - 5^(x + 3)\n\nDivide through by 3^(x + 1) :\n7 - 5 * (5/3)^(x + 1) = 27 - 25 * (5/3)^(x + 1)\n\nRearrange :\n20 * (5/3)^(x + 1) = 20\n\nTherefore, (5/3)^(x + 1) = 1 = (5/3)^0\nThus, x + 1 = 0, so, x = -1\n----------------------------------------------------------------------\nQuestion 3 : 6 * 9^(1/x) - 13 * 6^(1/x) + 6 * 4^(1/x) = 0\n\nFactorise for ease of viewing :\n6 * 3^(2/x) -13 * 2^(1/x) * 3^(1/x) + 6 * 2^(2/x) = 0\n\nDivide through by 6^(1/x), which equals 2^(1/x) * 3^(1/x) :\n6 * (3/2)^(1/x) - 13 + 6 * (2/3)^(1/x) = 0\n\nThe latter term, (2/3)^(1/x) = 1 / [ (3/2)^(1/x) ]\n\nSo now we have : 6 * (3/2)^(1/x) - 13 + 6 / [ (3/2)^(1/x) ] = 0\n\nLet y = (3/2)^(1/x)\n\nTherefore, 6y - 13 + 6 / y = 0\n\nMultiply through by y :\n6y^2 - 13y + 6 = 0\nor, (3y - 2)(2y - 3) = 0\nor, y = 2/3 or 3/2\n\nThat is, (3/2)^(1/x) = 2/3 or (3/2)^(1/x) = 3/2\n\nFirst, trying (3/2)^(1/x) = 2/3 = 1 / (3/2) = (3/2)^(-1)\nTherefore, 1 / x = -1, so x = -1.\n\nSecond, trying (3/2)^(1/x) = 3/2 = (3/2)^1\nTherefore, 1 / x = 1, so x = 1.\n\nSo there are 2 solutions, x = -1 or 1.
275079
1) m= (2-4)/(2-5) = -2/-3 = 2/3\n2) m= (1-1)/ (-8+9) = 0/1 = 0
729870
Answer 1:\n-5r = 55\nr = 55/-5\nr = -11\nChecking:\n-5r=55\n-5 (-11) = 55\n55 =55\n\nAnswer 2:\n2/8g=-14\n8g = -14/2\n8g = -7\ng = -7/8\nChecking:\n2/8g=-14\n2/8(-7/8) = -14\n2/-7 = -14\n-7 = -14/2\n-7 = -7
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
891080
1. Xc = 1/2πfC\nSo 2πfC = 1/Xc\nSo C = 1/2πfXc\n\n2. ib = ibq + √2is\nSo √ = ib - ibq\nSo 2is = (ib - ibq)²\nie is = ½(ib - ibq)²\n\n3. Q = XL/R\nSo XL = QR\n\n4. Lt = L1 + L2 - 2Lm\nSo 2Lm = L1 + L2 - Lt\nie Lm = ½(L1 + L2 - Lt)\n\n5. R1=i2 R2 / i2\n??????? So R1 = R2\n\n6. Es = E1 + E2 + E3\nSo E2 = Es - E1 - E3\n\n7. R1 / Rx = R2 / R3\nSo Rx / R1 = R3 / R2\nSo Rx = R1 * R3/R2\n\n8. y = Mx + b\nSo Mx = y - b\nSo M = (y - b)/x\n\n9. i2=i1 R1 / R2\nSo i1 R1 = i2R2\nie i1 = i2 R2 / R1\n\nI would be extremely concerned about your progress if you find this simple algebra a serious chore!!!!!! (Especially if you are learning the electrical trade)
789977
1.  15\n2.   a.  y=(2x-3)^2 + 4\n      b. (3/2;4)
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1) 4 3/4\n\n2) 1356.66\n\n3) 760
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1. x=1.25\nSol'n.\n1/3x+1/2=11/12\n1/3x=11/12-1/2\n1/3x=5/12\nx=15/12\nx=1.25\n\n2. x=-3\nSol'n.\n1.5x+9=4.5\n1.5x=4.5-9\n1.5x=-4.5\nx=-3\n\n3. x=28\nSol'n.\n2(x+2)=3(x-8)\n2x+4=3x-24\n24+4=3x-2x\nx=28\n\n4. x=-12\nSol'n.\nx+5=-7\nx=-7-5\nx=-12\n\n5. x=-10.5\nSol'n.\n-4/7x=6\nx=-42/4\nx=-10.5\n\n6. x=261\nSol'n.\nx/3+2=89\nx/3=89-2\nx/3=87\nx=261\n\n7. a=3.5\nSol'n.\n3a+13=9a-8\n13+8=9a-3a\n21=6a\na=21/6\na=3.5\n\n8. n=0.8\nSol'n.\n15n+9=21\n15n=21-9\n15n=12\nn=12/15\nn=0.8\n\n9. x=-1\nSol'n.\n2(x+6)=-2(x+-4)\n2x+12=-2x+8\n2x+2x=8-12\n4x=-4\nx=-1\n\n10.x=0\nSol'n.\n-(x+2)-2x=-2(x+1)\n-x-2-2x=-2x-2\n-3x+2x=-2+2\n-x=0\nx=0
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
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1. A & B cross  -  2\n2. A returns    -  1\n3. C & D cross  -  10\n4. B returns    -  2\n5. A & B cross  -  2\n\nTotal minutes: 17\n:)
502428
#1. (x-5)(x+1)\n#2. (x-3)(x+2)
164527
2 points to me for foolish questions!
437783
1) 1, 2, 3 and 6\n2) 1, 2, 2m\n3) 1\n4) 1, 3\n\n5) 5/9 (55/99) > (45/99) 5/11\n6) 2/3 (8/12) < (9/12) 3/4\n7) 4/5 (32/40) < (35/40) 7/8\n8) 1/3 (4/12) = (4/12) 4/12
987327
Find a number X such that  2X = (X/2) + 99\n\nmultiply both sides by 2 to get  4X = X + 198\nsubtract X from both sides to get 3X = 198\ndivide both sides by 3 to get X = 66\n\ncheck work\n2*66 = 132\n66/2 + 99 = 132