Any number inside the modulus sign becomes positive. This means 
and so we have,
Solving these gives us
However if we check the second solution in the original equation we obtain 
. This is false and so 
can't be a solution.
Therefore the only solution is 
.
(Note: I'm not sure why the second solution didn't work but when there's a modulus sign involved it always pays to check your final answers to be sure. I'll have a think about it but in case you find out before I do, I'd be interested to know in the comments.)
Answer:
Step-by-step explanation:
GIVEN: A farmer has 
of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is 
.
TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.
SOLUTION:
Let the length of rectangle be 
and 
perimeter of rectangular pen 
area of rectangular pen 
putting value of
to maximize 
but the dimensions must be lesser or equal to than that of barn.
therefore maximum length rectangular pen 
width of rectangular pen 
Maximum area of rectangular pen 
Hence maximum area of rectangular pen is 
and dimensions are
A radius of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length. Set up the formula for the area of a circle. The formula is A = π r 2 it equals the area of the circle, and r equals the radius.
Solve for the radius.
Plug the area into the formula.
Divide the area by.
Take the square root.
Just remember BODMAS, which B= brackets O= Order D= Division M= Multiply A=Addition and S=Subtract
So first you solve the brackets so 44+22 is 66
So 66-4+12 so 66-4 is 62
62+12 is 74
So now you have got 74 divided by 2 which is 37
Hope this helps xx
Y-65=3(x-49) 3 is time 49 is crushed
