how do you solve this using i's 7-i/1+i?
440770
i think u mean (7-i)/(1+i)\n\nto solve this u multiply the top and bottom by the complex conjugate of the denominator, so in this case by (1-i)\n\nso we get (7-i)/(1+i) * (1-i)/(1-i) = (7-8i+i^2)/(1-i^2)\n\nwe know i^2 = -1\n\nso answer is (6-8i)/2 = 3-4i
191203
(x-7i)(x+7i)=x^2 + (-i)^2 * 49 = x^2 + 49\n(which is your polinom having the roots -7i and +7i)
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
1099830
27\n\n(+7 -4 +8 -3 +9 -2 +10)
1078373
2z+7<z+10\nyou need to get like terms together so first...\n2z+7+-7<z+10-7\n2z<z+3\nnow get z over to the  2z\n2z-z<z-z+3\nz<3\n\ncheck ya later ♥
1054159
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
871212
i2=i1R1/R2\ni1=R2i2/R1\n\nR1/R2=i2/i1\n\nR1/Rx = R2/R3\n(R1/R2)*R3=Rx\nRx=(i2/i1)*R3
558121
i=i+2 (i=4)\ni++ (i=5)\ni=5
1169872
(7/(3x)) / (x/3)\n(7/(3x)) * (3/x)\n21/(3x^2)\n7/x^2\n\nBut if you meant\n\n((7/3)x)/(x/3)\n((7x)/3)*(3/x)\n(21x)/(3x)\nANS : 7\n\n-------------------------------\n\n2 1/3 - 3/4\n2 1/3 = (7/3)\n\n(7/3) - (3/4)\nMultiply everything by 12\n(28 - 9)/12\n(19/12) or 1 (7/12)
1007006
add or subtract 1\n\nor else subtract the S at the begining of Seven
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
934652
So let's say the problem looks like this\n\n   11\nx 12\n--------\n        <-----the answer from 11*2\n       <------answer from 11*1\n\nfirst hide the 1 from the 12 and work out the problem as 11*2\nthen work on multiplying the 1 with 11 but make sure you start putting your answer down below your first answer and the answer starts in the column that is below the 1 from the 12.\n\nfinally when the two answers are found, you can add them and get your final answer. SOOO here's what it looks like\n\n      11\n   x  12\n--------------\n           22    (this is from 11x2)\n +  11       (this is from 11 x1)\n------------\n     132 = final answer!
454578
mrniceguy's answer is correct, but he has one little eror in the work.  i^7 = -i
1089449
It's as simple as plugging in numbers.  The number 15 becomes the letter "a", the number 7 is "b", and the number 55 is "c".  Just substitute the number for the letter in the formula and solve.
968466
The key is that with division, you need to multiply the top and bottom by the conjugate of the denominator.  The conjugate is like this:\n\nThe conjugate of a+bi is\na-bi.\n\nIn this case, the denominator is 10+i.  The conjugate will be 10-i.\n\n(7+6i)/(10+i)=\n\nMultiply by the conjugate\n\n((7+6i)(10-i))/((10+i)(10-i))=\n\nSimplify, but remember that -i*i = -(-1) = 1.  So if you have an i and -i, the result will be positive. (treat i like a variable-use foil to expand)\n\n(70+53i+6)/(100+1)=\n\nSimplify\n\n(76+53i)/101\n\nYour real part will be 76/101, your imaginary part will be 53(i)/101
799993
Use the following to help solve this problem:\n\ni ^1 = i\ni ^2 = -1\ni ^3 = -i\ni ^4 = 1\n\nThis pattern then repeats:\n\ni ^5 = i^4 * i = 1 * i = i\ni ^6 = i^4 * i^2 = 1 * -1 = -1\n\netc.\n\nSo to solve any i ^n, just look at the remainder of n/4:\n\nremainder \nof n/4 ....... value of i ^n\n- - - -  - - - - - - - - - - - - - - -\n0 ....................... 1\n1 ........................ i\n2 ....................... -1\n3 ........................ -i\n\nSo now let's tackle your specific problem:\n\ni ^16 = 1 (since n=16, n/4 = 4 with remainder of 0)\ni ^6 = -1 (since n=6, n/4 = 1 with remainder of 2)\ni ^5 = i (since n=5, n/5 = 1 with remainder of 1)\ni ^13 = i (since n=13, n/4 = 3 with remainder of 1)\n\nSo, i^16 + i^6 - 2i^5 + i^13 = \n1 + (-1) - 2(i) + i =\n-i\n\nHope that helped!\n\n~ ♥ ~\n\nPS -- The "2-i" answers are incorrect.  Those answerers just made a tiny typo and took 1 - (-1) as the first two terms, rather than 1 + (-1).  Just wanted to let you know, so that you don't put the wrong answer down on your homework.
308039
[3x/(5x+15)][(7x+21)/3]\n= [3x/(5(x+3))][7(x+3)/3]\nCancel the 3's and the (x+3)'s\n= [x/5][7/1]\n= 7x/5