a+b+c+d=d+e+f+g=g+h+i=17;where a=4,find d&g?
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
878985
Now as you have it written, f(0) = 1, f(1) = 2, etc. so your equations are off.\nThey should read:\nf(1)=a+b+c+d...........=2... (1)\nf(2)=8a+4b+2c+d.....=4... (2)\nf(3)=27a+9b+3c+d...=8... (3)\nf(4)=64a+16b+4c+d.=15....(4)\n{we also know that f(0) = 1, so right away, we know that d=1.  But I wont use that now, just the four equations above. I didnt know your equations were off until I did most of the proof below}\n\n\nFirst, combine one equation with the other three to eliminate one variable.\nFor instance, subtract eq. (1) from the others to eliminate the d terms.\n\n8a+4b+2c+d =4\na+b+c+d = 2\nso 7a + 3b + c = 2\n\n27a+9b+3c+d = 8\na+b+c+d = 2\nso 26a + 8b + 2c = 6\n\n64a+16b+4c+d =15\na+b+c+d = 2\nso 63a + 15b + 3c = 13\n\nNow we have three equations and three unknowns:\n7a + 3b + c = 2\n26a + 8b + 2c = 6\n63a + 15b + 3c =13\nNow do the same thing.  Choose one of them and combine with the other two to eliminate another variable.  \nI'll choose 7a + 3b + c = 2\n\n26a + 8b + 2c = 6\n7a + 3b + c = 2\nMultiply the second one by 2 so that the c-terms are the same, then subtract\n26a + 8b + 2c = 6\n14a + 6b + 2c = 4\nso 12a + 2b = 2\n\n63a + 15b + 3c = 13\n7a + 3b + c = 2 \nMutiply the second one by 3 so that the c-terms are the same, then subtract\n63a + 15b + 3c = 13\n21a + 9b + 3c = 6\nso 42a + 6b = 7\n\nNow we have two equations and two unknowns.\n42a + 6b = 7\n12a + 2b = 2\nMultiply the second by 3 so that the b-terms are the same, then subtract from the first.\n42a + 6b = 7\n36a + 6b = 6\n6a = 1\na = 1/6\nNow that we know a, go back and substitute this value into one of the two equations with two unknowns to find b.\n36(1/6) + 6b = 6\n6 + 6b = 6\n6b = 0\nb = 0\nNow substitute these values of a and b into one of the equations with three unknowns to find c.\n7(1/6) + 3(0) + c = 2\n7/6 + c = 2\nc = 5/6\nThen substitute a,b,c into one of the original equations to find that\n d = 1.
555500
here are the answers:\n-17x-19 for the first one \n22x+1 for the second one\n-6x^2+3x+15 for the third one, which can also be written as -3(2x^2-x-5)
1181427
ABC+ABD+C'D\n=ABC+ABD(C+C')+C'D\n=ABC+ABCD+ABC'D+C'D\n=ABC(1+D)+C'D(AB+1)\n=1ABC+1C'D\n=ABC+C'D
448387
4b(5 - a) - 3c(5 - a)\n= (5 - a)(4b - 3c)\n\nThe answer is D. Since both have the factor (5 - a), you factor it out, and what's left is (4b - 3c).
924187
= 16ab/20ab^2\n= 4/5b // 16 and 20 can be reduced by 4, and an a and a b cancels out\n\n3*3*3*3*3 = 3^5\n3*3 = 3^2\n3^5/3^2 = 3^3\n= 3*3*3
240240
2/a + 8 - 1/5 +20\n\n27 and four fifths plus 2 over "a"
604743
It is 9 as the solution is 5+17+1+2 = 8+2+11+4 = 2+5+4+14 = 25\nTherefore 10+1+?+5=25\nTherefore ?=9\nTherefore the answers is 9.
766358
4(4)+2x7-(5x4-15)\n4x4\n16+2x7-(5x4-15)\n2x7\n16+14-(5x4-15)\n5x4-15\n16+14-5\n16+14\n30-5\n25=answer\n\n(b)-2bc+c(c)\n-2b^2c^3=answer\n\n2x-2-5x4/4+3x-1\n-4-20/4+-3\n-4-5+-3\n-2=answer
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
497634
To find the GCF of A and B:\n1. divide A by B and take the remainder.\n2. If there is no remainder, B is the GCF\n3. If there is no remainder, make B the new A and make the remainder the new B, and go to step 1.
150640
Since D= (L^2 + w^2 + H^2)^1/2 then\nD^2 = L^2 + W^2 + H^2\nW^2 =  D^2 - L^2 - H^2\nW = (D^2 - L^2 - H^2)^(1/2)\nOR W = SQRT (D^2 - L^2 - H^2)\nNOW FOR L = 12, D= 13 AND H=3 YOU HAVE\nW= SQRT (169 - 144 -9) = SQRT (16) = 4
81081
g(x)=(x-a)(x+a)/(x-a)\n= x+a\nSince g(x) =4\ntherefore, x+a = 4 \nbut, x=a so substituting the value of x\na+a=4\n2a=4\na=2\nThe constant a = 2.
305313
4x^3y^2-16y^3-9y=\ny(4x^3y-16y^2-9)
84658
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
721099
4a + 3b = 4x4 + 3x(-4) = 16 - 12 = 4
176413
Great puzzle, took me awhile to make it to f=1, g=4. The letters are from a-i: 9,8,7,2,6,1,4,5,3.\n\nPlugged in this is: (9 * 87 + 2/6) / (14 + 5/3) or (783 + 1/3) / (15 + 2/3) which equals 50.\n\nAnswer logged Sunday, March 12th, 3:20 pm with no other valid answers yet. Sorry to have to write that, but I've seen people change their answers too many times in the past.