62/18, 93/27, 124/36, 155/45,...To get the equivalent fractions of 3 4/9:1. Change the mixed fraction to improper fraction3 4/9:= [(9 * 3)+4]/9= [27+4]/9= 31/9 in improper fraction2. Multiply 31/9 by 2/2, 3/3, 4/4, 5/5,...31/9 * 2/2 = 62/1831/9 * 3/3 = 93/2731/9 * 4/4 = 124/3631/9 * 5/5 = 155/45Equivalent fractions of 3 4/9 = 62/18, 93/27, 124/36, 155/45,...
Yes, 599422 is divisible by 3. The sum of its digits is 5+9+9+4+2+2 = 31, which is divisible by 3.
9/4 x 31/9 = 7 3/4 or 7.75
31/9 = 3 and 4/9
31 ÷ 9 = 3 remainder 4
62/18, 93/27, 124/36, 155/45,...To get the equivalent fractions of 3 4/9:1. Change the mixed fraction to improper fraction3 4/9:= [(9 * 3)+4]/9= [27+4]/9= 31/9 in improper fraction2. Multiply 31/9 by 2/2, 3/3, 4/4, 5/5,...31/9 * 2/2 = 62/1831/9 * 3/3 = 93/2731/9 * 4/4 = 124/3631/9 * 5/5 = 155/45Equivalent fractions of 3 4/9 = 62/18, 93/27, 124/36, 155/45,...
Yes, 599422 is divisible by 3. The sum of its digits is 5+9+9+4+2+2 = 31, which is divisible by 3.
9/4 x 31/9 = 7 3/4 or 7.75
The method is... Step 1. Denominator x whole number Step 2. Add the numerator to that product Step 3. Keep the denominator and add in the sum of the numerator and the product as the numerator. Example: 3 4/9 1. 9x3=27 2. 27+4=31 3. 31=31/9 31/9 is the answer
x6 + 9= x6 - (-9) since i2 = -1= (x3)2 - 9i2 factor the difference of two squares= (x3 + 3i)(x3 - 3i) since 3 = (31/3)3 and -i = i3 we can write:= [x3 - (31/3)3i3] [x3 + (31/3)3i3]= [x3 - (31/3i)3] [x3 + (31/3i)3] factor the sum and the difference of two cubes= [(x - 31/3i)(x2 + 31/3ix + (31/3)2i2)] [(x + 31/3i)(x2 - 31/3ix + (31/3)2i2)]= [(x - 31/3i)(x2 + 31/3ix - (31/3)2)][(x + 31/3i)(x2 - 31/3ix - (31/3)2)]Thus, we have two factors (x - 31/3i) and (x + 31/3i),so let's find four othersAdd and subtract x2/4 to both trinomials[x2 - x2/4 + (x/2)2 + 31/3ix - (31/3)2] [x2 - x2/4 + (x/2)2 - 31/3ix - (32/3)2] combine and factor -1= {3x2/4 - [((x/2)i))2 - 31/3ix + (31/3)2]}{3x2/4 - [((x/2)i))2 + 31/3ix + (32/3)2]} write the difference of the two squares= {((3)1/2x/2))2 - [(x/2)i - 31/3]2}{((3)1/2x/2))2 - [(x/2)i + 32/3]2]} factor the difference of two squares= {[(31/2/2)x - ((1/2)i)x - 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - (((1/2)i)x + 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]}= {[(31/2/2)x - ((1/2)i)x + 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - ((1/2)i)x - 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]} simplify= {[((31/2 - i)/2))x + 31/3)] [((31/2 + i)/2))x - 31/3)]} {[((31/2 - i)/2))x - 31/3)] [((31/2+ i)/2))x + 31/3)]}so we have the 6 linear factors of x2 + 9.1) (x - 31/3i)2) (x + 31/3i)3) [((31/2 - i)/2))x + 31/3)]4) [((31/2 + i)/2))x - 31/3)]5) [((31/2 - i)/2))x - 31/3)]6) [((31/2+ i)/2))x + 31/3)]Check: Multiply:[(1)(2)][(3)(5)][(4)(6)]A) (x - 31/3i)(x + 31/3i) = x +(31/3)2B) [((31/2 - i)/2))x + 31/3)] [((31/2 - i)/2))x - 31/3)] = [(1 - (31/2)i)/2]x2 - (31/3)2C) [((31/2 + i)/2))x - 31/3)][((31/2+ i)/2))x + 31/3)] = [(1 + (31/2)i)/2]x2 - (31/3)2Multiply B) and C) and you'll get x4 - (31/3)2x2 + (31/3)4Now you have:[x +(31/3)2][x4 - (31/3)2x2 + (31/3)4] = x6 + 9
They are 1*1 = 1 2*2 = 4 3*3 = 9 and so on until you get to 31*31 = 961
31/9 = 3 and 4/9
31 ÷ 9 = 3 remainder 4
9*9/3 + 6 - 2 = 9*3 + 6 - 2 = 27 + 6 - 2 = 31
Let the fractions be x and y:- If: 2x+3y = 31 Then: 2/31x+3/31y = 1 by dividing all terms by 31 So: 2/31*31/4 + 3/31*31/6 = 1 because 1/2+1/2 = 1 Therefore: x = 31/4 and y = 31/6 Check: 2*31/4 + 3*31/6 = 31
3-3, 3-3, 3-3, 3-3, 4-1, 2-1, 2-3
3 squared times 2 times 10 32*2*10 = 9*2*10 = 180