Exactly 31, strangely enough!
You can invent infinitely many expressions that have a value of 31, for example: 31 31 + 0 31 + 0 + 0 30 + 1 30 + 1 + 1 1 x 31 5 x 6 + 1 etc.
3 and 31 because 9+3=12 which is 3 because it goes into 12 evenly and 31 because 93 divided by 31 is 3 or 3 times 31 is 93
1 x 31, 2 x 31/2 , 3 x 31/3 etc. 31 is a prime number.
1/3 + 1/31/3 + 1/31/3 + 1/31/3 + 1/3
31 = 3-4k 31-3 = -4k 28 = -4k Divide both sides of the equation by -4 to find the value of k: k = -7
The 3 is in the ten thousands' place so its vaue is thirty thousand.
4x + 3, where x = 7 4(7) + 3 = 28 + 3 = 31
The absolute value is a number's distance from zero on a number line. Therefore the absolute value of -31 is 31.
Exactly 31, strangely enough!
31 + 216 = 247
A percent means a hundredth part - Thus:- Divide 354 by 100. This gives you the value for 1%. Then multiply the 1% value by 31 to give you the value of 31%. (345/100)*31 = 106.95
3% of 31 = 31*3/100 = 3/3100
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
of 31 = 100*3/31 = 9.677 % approx.
You can invent infinitely many expressions that have a value of 31, for example: 31 31 + 0 31 + 0 + 0 30 + 1 30 + 1 + 1 1 x 31 5 x 6 + 1 etc.
-3