_//(2*2) - 4x - 7// * _//2// = 0 4* 12x - 7 * 122/100 = 0 28*12x * ((28*12)/122) = 0 (23/10 * x) * ((23/10 * x)/122) = 0 23/10 *x = 0 / ((23/10 *x)122) 23/10 *x = 0 23/x = 10 x = 2.3
X + Y = 23 XY = 132 Y = 23-x xy = x (23-x) = 23x -x2= 132 x2-23x+132 = 0 (X-11)(x-12) = 0 X = 11 Y = 12
187
xy=24 and y=23-x so x(23-x)=24, which is x2-23x+24 = 0, and you solve this using the formula method to get approximately 21.95 and 1.05 If the product had been 23 and the sum 24 the equation would have factorized easily and you would have 1 and 23
1011 = 1 x 23 + 0 x 22 + 1 x 2 + 1 x 1 = 11
(x^2)-23x=0 Start by factoring out an x: (x)(x)-23x=0 x(x-23)=0 Set each quantity to 0 and solve for x: x(x-23)=0 x=0 x-23=0 x-23+23=0+23 x=23 Therefore x=0, x=23
x=0
_//(2*2) - 4x - 7// * _//2// = 0 4* 12x - 7 * 122/100 = 0 28*12x * ((28*12)/122) = 0 (23/10 * x) * ((23/10 * x)/122) = 0 23/10 *x = 0 / ((23/10 *x)122) 23/10 *x = 0 23/x = 10 x = 2.3
(23/2)2 x2 - 23x = 0 x2 - 23x + (23/2)2 = (23/2)2 (x - 23/2)2 = (23/2)2 x - 23/2 = 23/2 x = 23/2 + 23/2 x = 23 Of course you could come to this solution much more easily by dividing the whole equation by x at the start. There really is no need to complete the square in this case: x2 - 23x = 0 x - 23 = 0 x = 23
To get the multiples of 23, just multiply 23 x 0, 23 x 1, 23 x 2, etc. - Or, starting with zero, add 23 at a time.
X + Y = 23 XY = 132 Y = 23-x xy = x (23-x) = 23x -x2= 132 x2-23x+132 = 0 (X-11)(x-12) = 0 X = 11 Y = 12
no, eg. 409 x 2 = 818 19 x 0 + 23 = 23
similar to a base 10 number... Ex: 100100 contains 6 place values. The left most is 1 x 26-1 , the second is 0 x 25-1 , and 0 x 24-1 ,1 x 23-1 ,0 x 22-1 ,0 x 21-1 . or 1 x 25 , 0 x 24 , 0 x 23 , 1 x 22 , 0 x 21 , 1 .
25
It is zero because anything multiplied by 0 is zero
f : x -> 3x + 2 where 0 < x <23.
3x+8-31=0 Simplify it. 3x-23=0 Add 23 to both sides to get rid of the -23 3x=23 Divide both sides by 3 x=23/3 Simplfy the fraction x=9 2/3 So there we have it, x is nine and two thirds (or 9.6666...)