Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.
Substitute the known values into the formula:
x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}
Simplify:
<span>x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}</span>
<span>x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]</span>
<span>x√10=√[2x(x)(x-1)(1)]</span>
<span>x√10=√[2x²(x-1)]</span>
<span>x√10=√(2x³-2x²)</span>
<span>10x²=2x³-2x²</span>
<span>2x³-12x²=0</span>
<span>2x²(x-6)=0</span>
<span>2x²=0 or x-6=0</span>
<span>x=0 or x=6</span>
<span>Therefore, x=6 (you can't have a length of 0).</span>
Hello!
Brice is standing further away from the door, because 20 metres are greater than 20 centimetres.
A lot greater, actually, because 20 metres = 2,000 centimetres.
H: 5 square yards is the correct answer, break the figure into squares and then multiply length by width to get area
Y = -3x + 6
y - 6 = -3x
-1/3 y + 2 = x
z = x -3
z = -1/3 y + 2 -3 = -1/3 y - 1
set z = 0
-1/3 y = 1
y = -3
set y = 0
z = - 1