Answer:if it’s a pretest just guess you don’t need to get it right
Step-by-step explanation:
9514 1404 393
Answer:
- square: 12 ft sides
- octagon: 6 ft sides
Step-by-step explanation:
This problem can be worked in your head.
If the perimeters of the square and regular octagon are the same, the side length of the 4-sided square must be the same as the length of 2 sides of the 8-sided octagon. Since the side of the square is 6 ft more than the side of the octagon, each side of the octagon must be 6 ft, and each side of the square must be 12 ft.
__
We can let s represent the side length of the octagon. Then we have ...
8s = perimeter of octagon
4(s +6) = perimeter of square
These are equal, so ...
4(s +6) = 8s
s +6 = 2s . . . . . . divide by 4
6 = s . . . . . . . . . . subtract s
The octagon has 6-ft sides; the square has 12-ft sides.
Answer:
yes, i think it is correct :)
Step-by-step explanation:
Answer:
512
Step-by-step explanation:
Suppose we ask how many subsets of {1,2,3,4,5} add up to a number ≥8. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to 15. Exactly one of those parts is therefore ≥8. There must be at least one such part, because of the pigeonhole principle (specifically, two 7's are sufficient only to add up to 14). And if one part has sum ≥8, the other part—its complement—must have sum ≤15−8=7
.
For instance, if I divide the set into parts {1,2,4}
and {3,5}, the first part adds up to 7, and its complement adds up to 8
.
Once one makes that observation, the rest of the proof is straightforward. There are 25=32
different subsets of this set (including itself and the empty set). For each one, either its sum, or its complement's sum (but not both), must be ≥8. Since exactly half of the subsets have sum ≥8, the number of such subsets is 32/2, or 16.
Well If your trying to turn it into a decimal the answer would be .25 but if you want it as a fraction the answer is 100 over 25. Hope I helped!
-Amber
