ALGEBRA II question....can any1 answer dis one?
970283
Your basic equation for volume is \nV = Pi * R^2 * H        (this is the answer to part B)\nIf you put the values for radius and hight into that equation you have\nV = Pi * 4 * 6\nWe know Pi is 3.14159, so the equation solves\nV = 75.39822369\nI think you mistyped the 6 when aiming at the 9.\n\nYour constant is Pi.\n\nPutting the values for R=3 and H = 5 into that equation will get \nV = Pi * 3^2 * 5\nV = 3.14159 * 9 * 5\nV = 141.372
1089643
http://www.algebra.com
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped
223613
dont the value of x have to be the same? to the person above me..\n\ni thought that x^2 means x to the power of 2 (x squared)\nooo wait i dont know im confusing myself. im only 13 and not that smart but these are my answers now.\n\n\nfirst i got x= negetive eight.\nthen i got x= 2 and -8\n\n\nand gary, wow thats befuddling.\nyour a nerd.\n\nwhat are you kidding me?? thats so not seventh grade math! thats algebra, and im only in pre algrebra.
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
556722
look at my question\n\nBidDogARRRRR...\n1 minute ago\nDoes anyone know 0=1 and anything = anything else?\na = 1; b = 0\n1. a = b + 1 \n2. (a-b)a = (a-b)(b+1)\n3. a2 - ab = ab + a - b2 - b \n4. a2 - ab -a = ab + a -a - b2 - b\n5. a(a - b - 1) = b(a - b - 1) \n6. a = b\n7. b + 1 = b \n8. Therefore, 1 = 0
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
395828
haha, but this does not work. you cannot mathematically divide by (a-b) because this quantity is equal to 0. (1-1)=0. Dividing by zero yields an indeterminate answer.
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
1036177
sory i didnt get ur Q\ndo u mean 2x-6>x ?\nthen x is greater than 6
571127
Try going to this website. The will explain it step by step and you can ask all the math questions you want. www.algebra.com
1022192
i have  a book about algebra..... and some website...see if they could help u
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
839872
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
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Is the = really a minus?  If so, the GCF of each term is\n4a^2b, because the GCF of 4 and 8 is 4, and the GCF of variables has the lower of the exponents of each variable.\n\nDivide each term by this GCF (when you divide, subtract the exponents) and write the GCF outside () with the division answer inside (distributive property in reverse)\n\n4a^2b(b^2 - 2ac)