Let
x---------> length original square side
x1--------> length resulting rectangle
y1--------> wide resulting rectangle
A1--------> area resulting rectangle
we know that
x1=x+2
y1=x-2
A1=60 in²
60=(x+2)*(x-2)--------> 60=x²-2x+2x-4-------> 60=x²-4------> x²=64
x1=8
x2=-8
the solution is
x=8 in
<span>the area of the original square is 8*8-------> 64 in</span>²
the answer is 64 in²
if y-6 > 10, and 16-6 = 10, then y > 16 since 17-6 = 11, which is greater than 10. Any number greater than 16 can fulfil this inequality.
Answer:
what do you need help with ^_^
Step-by-step explanation:
Answer:
6a^3+ 22a^4+ 14a-10
Step-by-step explanation:
6a^3 10a^2 12a^2 20a -6a -10
2x^2-5x-3
This is the answer I believe, I apologize if it’s incorrect
