I think it’s not a different of squares because when u factor it =2(y-8)
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>
Here are the log properties you need:
Step-by-step explanation:
i.1+y<25
y<25-1
y<24
ii.101+6y>200
6y>200−101
6y>99
y>99/6
y>33/2
