In using the rules of exponents, simplify the following espression...?
919306
3m^1/6 n^1/3 over 4n^-2/3 all to ^2\n= ( 3m^1/6 n^1/3 / 4n^-2/3 )^2\n= (3 m^1/6 n^{1/3+2/3} /4 )^2\n= (9/16) m^{2/6} n^2\n= (9/16) m^{1/3} n^2 '
926880
(1+1)-2*3/2^2+2\n\nWithout useing Order of Operations: 2\n\nWith Order of Operations: 2.5
966462
(6x^2)/(4x^2 y-12xy) * (x^2 + x - 12)/ (3x^2 + 12x)\n\nFirst, I would attempt to simplify if possible:\n2x(3x)/x(2xy-6y) * (x^2 + x - 12)/ (3x^2 + 12x )\n(3x) / (2xy - 6y) * (x^2 + x - 12)/ (3x^2 + 12x)\n\nRewrite to see if we can cross cancel:\n\n(3x)           (x^2 + x - 12)\n---------  *   -----------------\n(2xy-6y)    (3x)(x + 4x)\n\n1               (x^2 + x - 12)\n---------- *  -----------------\n(2xy-6y)     (x+4x)\n\nUnfoil the (x^2 + x -12 to see if we can further simplify\n\n(x^2 + x - 12) = (x -3)(x+4) so you will have:\n\n1             (x-3)(x+4)\n-------  *  --------------\n2y(x-3)      (x+4)\n\nBy simplification:\n\n1           (x-3)\n------   * ----------\n2y(x-3)     1\n\nBy Cross Cancel:\n1/2y
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
851770
take the log\nso log b= log (1+(r/n))^n*t\nlog b = nt log (1+(r/n))\n{(log b)} / {log (1+(r/n))} = nt\n[{(log b)} / {log (1+(r/n))}] / n = t\ni hope those parenthesis dont mess u up lol\ni just learned this in school
783096
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
880382
Ok, I think that I got it. on the third one you posted, it's about simplifying the binomials (2 numbers). So you add the exponents of the like variables (or coefficiants- there's no number, then it's assumed to be 1. Ex.\n(z^4)^-3 * (z^-2)^-5\n(z^2)^-8\n\nAre there no answers in the back of the book? I would email you, but it blocks your address, if I could see some of your notes it would be easier to explain I think.
830308
Denominator can be written as (x^2-a^2).(x^2+a^2)\n\nThe x^2-a^2 terms cancels when a is not equal to zero leaving the formula as\n\n1/(x^2+a^2)\n\nThe denominator here approaches zero as x approaches a \nso the original formula tends to infinity.  Note that since squares are positive the answer is positive infinity as x approaches a from the left and the right.\n\nOther ways are to use L'Hopitals rule.
381470
You add exponents when you are multiplying similar base numbers raised by exponents.  So, 6 to the third power times 6 to the second power is 6 to the fifth power. The rule is important when dealing with expressions and equations with variables, especially multiple variables.  Good luck.
514501
It's not really a problem as much as a simplification; there's no = sign.  Anyway, simplified I get 8w^3 + 4w^2.  I got that by multiplying your original problem by \n(-2w-1)/(-2w-1).  That gets rid of the fraction in the denominator and reduces the numerals in the numerator down.
769880
The exponent is not divided by (1 + 1) in 2*x^(1+1)/(1+1), the whole expression is. ie 2x²/2 = x². Its not to be confused with 2x^(2/2) ie 2x^1 = 2x\n\nWhen there are limits it makes no difference\n\neg ∫(from x = a to x = b) 2x dx \n= [x² + c](from x = a to x = b)\n= (b² + c) - (a² + c)\n=  b² - a² (irrespective of the value of c)
731776
simple!!!\nthe answer is : 12x^2\nthe rule of addition intergers with exponents is simple.\nyou just add the whole number (that is 4+8) and then follow the integers (that is x^2)... get my point?\ni hope i have helped you with your assignment....
1010071
ok its like this...() or parenthesis meant multiplication. Then multiplyin negative by negative is = to +. So now its gonna be +a or simply a
735282
Seems like you are missing some parentheses and an exponent. Please check to make sure yiu have copied the problem correctly and resubmit.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
56508
first, do 2 to the 12th power (exponents take precedent over multiplication) which is 4096. Then multiply that by 86 and get 352256.  It's quite simple, really.
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.