Answer:
( 8 , - 8) => x² - 55 = 9
( 4 , - 4) => 2x² - 32 = 0
(5 , - 5) => 4x² - 100 = 0
(11 , - 11) => x² - 140 = -19
( 3, - 3) => 2x² - 18 = 0
Step-by-step explanation:
1)

x = ( 4 , - 4)
2)

x = ( 5 , - 5 )
3)

x= ( 8 , - 8)
4)

x = ( 11 , -11)
5)

x = ( 3 , - 3)
Answer: 160mi
Step-by-step explanation:
To find the area of a rectangle or that kind of shape u need to find the height and width and that would be 8mi and 20mi and of course they're the same on both side so u need to Height x Width so 8x20=160 or 20+20+20+20+20+20+20+20=160 or 8+8+8+8+8+8+8+8+8+8+8+8+8+8+8+8+8+8+8+8=160 but nobody wants to do it that way unless they absolutely have too
Answer:
40% is the answer to your question.
The last one is rational because it is a terminating decimal.
Answer:
Check photo
Step-by-step explanation: