I dont know how ot make exponent sing but its x with a 2 above it. plz help TY
130751
are you trying to solve for x?\n\nFirst, find the factors for the equation.  (x+2)(x+1)\n\nCheck it to see if it's right.  x times x equals x squared. 1x plus 2x equals 3x. 2 times 1 equals 2.\n\nMake one of the parenthesis groups equal to zero.  For example, x+2 = 0   To solve this, x must equal -2.\n\nPut -2 back into the original equation and see if it works.\nNow make the parenthesis group equal to zero.  x+1=0  To solve this, x must equal -1.\n\nTherefore, the answer is using either -1 or -2 will make the equation zero.\n\nHope this helps.  ;)
590052
1) when you are multiplying them together, just add the exponents.  m^5\n\n2) x^2 / x^5 is the same as (xx) / (xxxxx), so simplify it by removing 2 x's from each side, and you have 1 / (xxx) which equals 1/x^3\n\n3) 0.00317 = 3.17 x 10^-3  to do a problem like this just count how many places you have to move the decimal to get it with only one number on the left.  If you have to move it to the right, then the answer is 10^ negative how many places you counted.  If you move it to the left, it's 10^ positive.\n\n4)\n2.3(10 ^ -2) times 4.1(10^-3)\n(2.3)(4.1)(10^-2)(10^-3)\n(2.3)(4.1)(10^-5)\n(9.43)(10^-5)\n9.43 x 10^-5\n\n5) hmmm,  it's been a really long time, but i believe the degree of the equation is the highest exponent, so 4th degree?\n\n6) first multiply every number inside the first parenthesis together with the numbers in the second set of parenthesis, then simplify. \n(3y^2 - 7y + 4) (2y^2 + 3y - 2)\n\n(3y^2)(2y^2) + (3y^2)(3y) + (3y^2)(-2) + (-7y)(2y^2) + (-7y)(3y) + (-7y)(-2) +(4)(2y^2) + (4)(3y) + (4)(-2)\n\n6y^4 + 9y^3 - 6y^2 - 14y^3 - 21y^2 + 14y + 8y^2 + 12y - 8\n\n(6y^4) + (9y^3 - 14y^3) - (6y^2 - 21y^2 + 8y^2) + (14y + 12y) - 8\n\n6y^4 - 5y^3 - 19y^2 +26y - 8\n\nuse that same principle to do 7 and 8, they are the same line of thought, just easier.\n\nthat's all i've got, i should be working, not doing algebra.
450835
if you are using a "while" or "do while" loop\n\nwhile ((x1 - x0) > 0.001)\n{\n//your code here\n}\n\nor you can do this\n\ndo\n{\n//your code here\n}\nwhile ((x1 - x0) > 0.001);
361182
If I understand you, the exponents are 3x+4, x, and 12, all to base 8. In multiplying the left hand side, add the exponents:\n(8 to the 3x+4) times 8 to the x equals 8 to the (3x plus 4 plus x).\nSo, 8 to the 4x plus 4 equals 8 to the 12.\n4x plus 4 equals 12\nx equals 2
620475
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
206019
x=-(-1)+/_ {(-1)^2 -4*5*12} all over 2*5\n= 1+/- {1-240} over 10\n= 1+/- i{249} over 10\n\n{}= square root, ^= to the power of,\n\nif you need more help let me know via email.
57476
i'm good at math, but what exactly is the quustion??? there isn't any... what is it, a equation?\nany way you can post it to sciences and mathematics > mathematics, im sure someone there can help you...\n\n>>>ok one minute i'm working on it...\n\n>>>there you go:\nlets do the first: (x-3)(x-1)(x+2)\nfirst we open the last two parentheses as in usual opening, and the (x-3) remains as it is. we get (x-3)[x^2+2x-x-2]\nequals (x-3)(x^2+x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-3). then we get:\nx*x^2+x*x+x*(-2)-3*x^2-3*x-3*(-2)\nx^3 + x^2 - 2x -3x^2 -3x + 6\nand finally:\nx^3 -2x^2 -5x + 6\n\nthe second is the same way, i'm doing it, good luck :)\n\n>>>the second: (x-5)(x+1)(x-2)\nfirst we open the last two parentheses as in usual opening, and the (x-5) remains as it is. we get (x-5)[x^2-2x+x-2]\nequals (x-5)(x^2-x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-5). then we get:\nx*x^2+x*(-x)+x*(-2)-5*x^2-5*(-x)-5*(-2)\nx^3 -x^2 -2x -5x^2 +5x +10\nand finally:\nx^3 -6x^2 +3x + 10\n\ni hope i helped...\n\n>>>oh i leave the first one to the last because i have to leave one of the three, and it really doesn't matter which one i leave. for example on the first i left the first parenthese (x-3)[(x-1)(x+2)] but i could leave the last [(x-3)(x-1)](x+2) or the middle (x-1)[(x-3)(x+2)] and i'd still get the same answer. i have to leave one of the three bcuz i can't open them all together, it's complicated and hard, and by opening two ane leave one it becomes much easier.
521929
it's easy of means multiply something the .7  is in precent and u have to convert it into a decimal by moving the decimal point 2 spaces to the left. the answer to the problem is.9093. I think u are missing some infor u probably want a fraction like is/for = %/whole I can't do that because u have not provide me with details.
20804
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
559020
u r probably working on order of operations, this might help:\nPlease (parantheses)\nExcuse (exponents)\nMy  (multiplication)\nDear (division)\nAunt (addition)\nSally (subtraction)\n\n1. 7000 x 400000 = 2800000000\n2.  1/144 = x^2 so x = +/- 1/12\n3.  -3 +/- the square root of 2\n4.  5t^4 - 7t^3 + 6t -2\n5.  4x^4 - 16x^2 +16
437315
Let's simplify what is in the parantheses first.  You can cancel out some of the x's and y's.\n\n(y^2/x^4)\n\nI canceled out the x and subtracted one from 5 (x^5) and I canceled out the y on the bottom and subtracted one from 3 (y^3)\n\nTo use a negative exponent, you must flip the fraction.  So\n\n(y^2/x^4)^-2= (x^4/y^2)^2\n\nNow square each term, which means multiply the exponent by 2.\n\nx^8/y^4\n\nYou were correct!  Hope this helps.
482020
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
417398
x³y - 3x²y - 10xy --- Take x and y out of each term...\nxy(x² -3x - 10) --- Factor the x² - 3x - 10\nxy(x - 5)(x + 2)\n\n*Note: To show exponents, such as x², use the ^ key (such as x^2). This expression looks like you basiclly wrote:\n3xy - 6xy - 10xy
443985
I'll rewrite your equation so everyone understands it:\n( ^ is short for "raised to" or exponent )\nFor example: y * y * y = y^3\n\n(3x^5y)^2\n\nThis is the same as (3x^5y) (3x^5y), right?\n\nYour answer is correct:\n9x^10y^2.\n\nGood luck in your studies.
531173
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
495742
I assume that the last ex you wrote is e^x too, right?\ny' =  e^(e^x) + x [e^(e^x)] (e^x)\ny' = e^(e^x) [1+ x e^x]\n\nY''= e^(e^x)[ xe^x + e^x] + e^(e^x)  e^x [1 + xe^x]\ny''= e^(e^x)[  xe^x + e^x + 1 + xe^x]
139309
taylor expansion of tan^2(5X) and ignore any term with x's power great than x^2. this should give you 25x^2. and divide that by 10x^2 gives you 2.5. and sing your happy song with a happy face.