Arectangle has a length at most two more than seven times the width. if the perimeter is less than or equal to 52 units, find the lest possible width and length.

expression #3 is not equivalent. the others are.

answer: x=60

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~