i need help. (7x^3y^-5)(4xy^3) Simplifing that problem.Its Adv Alg. hmw question.?
927974
Multiply the coefficients and then add and subtract the exponents for the like bases. \n(7x^3y^-5)(4xy^3)= 28x^(3+1)y^(-5+3)= 28x^4y^-2
904822
Hello Dear Blue \n\n2x^3+8x^2 - 5x + 5 ÷ x-2\n\n▪ Step1\n( 2x^2 ) ( x-2 ) = 2x^3 - 4x^2 ; { let Q1 = 2x^2 }\n2x^3+8x^2 - 5x + 5 - ( 2x^3 - 4x^2 ) = \n2x^3+8x^2 - 5x + 5 - 2x^3 + 4x^2 = +12x^2 - 5x + 5 {remainder} \n\n▪ Step 2\n+12x^2 - 5x + 5 ÷ x - 2 \n(x - 2) (12x) = 12x^2 - 24x ; { let Q2 = +12x }\n+12x^2 - 5x + 5 - ( 12x^2 - 24x  ) =\n+12x^2 - 5x + 5 - 12x^2 + 24x = +19x +5 {remainder}\n\n▪ Step 3 \n19x +5 ÷ x -2\n(x -2) (+19) = +19x - 38; { let Q3 = +19}\n19x +5  - (+19x - 38) =\n19x +5  -19x + 38 = +43 {remainder}\n\n▪ Step 4\nSum Q1, Q2, Q3\nQ1 + Q2 + Q3\n2x^2 +12x +19\n\nSo;\n(2x^3+8x^2-5x+5) ÷ (x - 2)    = >\n\nQ(x) OR quotient = 2x^2 +12x +19 \nR(x) OR Remainder is + 43 \n\nGood Luck dear.
235893
(15y^3 + 12y^2)+ (5y+4)\nGCF 3y^2(5y+4)+1(5y+4)\n\n(5y+4)(3y^2+1)
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
868484
-12x^2*3y / -21xy^4\n= -36x^2 y / -21xy^4\n= 12x / 7y^3
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
445755
are you simplifying? if so, here's your answer:\n\n(x^2-5x+6)(x^2-5x+5)=\nx^4-5x^3+5x^2-5x^3+25x^2-25x+6x^2-30x+30=\nx^4-10x^3+36x^2-55x+30\n\nwtf cranger, i got it right! what do you mean by CORRECT? check the bottom line
843483
using (x+y)^3 = x^3+3x^2y+3xy^2+y^3\n\nwe get (2a+b)^3 = 8a^3+12a^2b+6ab+b^3\n\n(ab+c)^3 = a^3b^3+3a^2b^2c+3abc^2 + c^3\n\n(6ab-c)^3 = 216a^3b^3-108a^2b^2c+18 abc^2 - c^3
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
491746
If u could give the entire question i could solve it for u and i'm not jokin. its actually quite easy.Its basically\n\n(w+3)/4w * 2w/(w-3).So wat ud have is\n\n1/2( (w+3)/(w-3) )\n\nits xactly like sayin 2/4 = (2/1) / (4/1)
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.
774020
3x^2(5x^2 – 2x + 6) – 2x(x^2 + 3x + 9)\n15x^4-6x^3+18x^2-2x^3-6x^2-18x\n15x^4-8x^3+12x^2-18x\nx(15x^3-8x^2+12x-18)
810076
11^(0.35+7) = 3 this is a contradiction\n\nx=3 is a solution but the rest is just a mess.\n\nDo you have any idea of what you are doing?
728357
3x^2y^2(2y+3y)/3x^2y^2\n=5y
947283
You can do this using "synthetic division"\n....... ______________________\nx - 5 ) x^4 - 0x^3 - 3x^2 + 0x + 6\n\nI'm going to simplify this leaving out the x-terms and only showing the coefficients, but just imagine they are there:\n\n.............. 1 . 5 .22 110\n1 -5 ) 1 . 0 .-3 . 0 . 6\n......... 1. -5\n......... ------\n.............. 5 .-3\n.............. 5 -25\n.............. -------\n................... 22. 0\n....... ........... 22 -110\n....... ............----------\n....... ........ ........ 110  6\n....... ........ ........ 110 -550\n....... ........ ........ ------------\n....... ........ ........ ....... 556\n\nSo if I did it correctly the answer should be:\n1 5 22 110 remainder 556\n\nThis equates to:\n(x^3 + 5x^2 + 22x + 110) + 556/(x-5)