CPhill

I assume that you want to find the least common multiple. This is easy...let's factor each number completely.....

20 = 2² x 5

22 = 11 x 2

23 = 23 x 1

Now...take each different factor and multiply them together...if we have two factors (or more) that are the same, choose the one with the highest power. (exponent)...so we have

2² x 5 x 11 x 23  = 5060

And that's the least common multiple.

(1 Å * = 10^-10 m )

(1m = 10^-3 km  )   

So 1 Å =10^-13 km

Angle ACB = 1/2 of its intercepted arc. And AB is the arc it intercepts. Thus ACB = (1/2)128 = 64 degrees. And BAC is the other angle in triangle ABC, so it equals 180 - 86 - 64 = 30 degrees.

OK, DS......we have a triangle ABC where one angle  = 110 degrees. Note that two of the sides  are just radii of the circle. And all radii in a circle are equal.  And by a theorem in Gepmetry, when two sides of a triangle are equal, the angles opposite those sides (in this case, angle A and angle B) are equal. So remember , that in any triangle, the angles always sum to 180 degrees. So angle A + angle B + 110 = 180. But angle A = angle B, so we can write......angle A + angle A + 110 = 180. Therefore 2*Angle A + 110 = 180. Subtract 110 from each side, and we have....2*Angle A = 70. Now, divide both sides by 2, and we have that Angle A = 35.

Got it???

The two radii are equal sides of an isosceles triangle. So the angles opposite those sides (A and B) are equal. Thus, angle A + angle B + 110 = 180. So, angle A + angle A + 110 = 180. Subtract 110 from each side. Thus angle A + angle A = 180-110. So 2*angle A  = 70. Divide by 2 on both sides. Then, angle A = (70)/2 = 35 degrees.

Simplifying, we have...

x - 6 - 6x + 36 = 0 

-5x + 30 = 0     subtract 30 from both sides

-5x = -30     divide both sides by -5

x = 6

Very nice explanation, Aziz...!!!

Here is another way to look at this, analytically (if not particularly "mathematically")...Note that, when lyl =0,  lxl cannot be larger than 1. Thus, x cannot be more than +1 or less than -1. So, plot the two points on the graph (1,0) and (-1,0). Likewise, when lxl = 0, lyl cannot be more than 1 or less than -1. Then, also plot the two points (0,1) and (0, -1) on the graph. Then, let all four points be the vertices of a square. The sides of this square are the "bounds" of our graph. Note that every point lying on any of the four sides satisfies lxl + lyl = 1, and any point lying "outside" these bounds, doesn't. (They make the equation > 1).

This gives us a way to graph any equation of the form lxl + lyl = a....where a is some positive real number.

Does that help??

The area is a square with a side of  √2....thus the area = ( √2)²  = 2

A graph might help........

Key it in like this......

.105-1.96*sqrt(.105*.895/258)