Answer:
The little blue-gray critters that live under rocks and logs and roll into a ball when disturbed go by many names
Explanation:
The little blue-gray critters that live under rocks and logs and roll into a ball when disturbed go by many names: roly-poly bugs, pillbugs, woodlice, tiggy-hogs, parson-pigs and their scientific name, Armadillidium vulgare. Contrary to popular belief, roly-polies are not even technically bugs. But they do play an important role in ecosystems
Answer:
At the end of 6 hours, 6 feet of the ladder remained above water
Explanation:
we are given the following:
change in tide = 8 inches per hour
1 hour = 8 inches
∴ 6 hours = 8 × 6 = 48 inches
Now, since the length of the ladder is measured in foot, let us convert the rate of tide rising to foot.
12 inches = 1 foot
dividing both sides by 12:
Therefore, after 6 hours, the tide rises by 2 feet.
Next, we are told that the rope ladder at the beginning was 8 feet, hence the length remaining above water after 6 hours of the rising of the tide is calculated as follows:
= 8 feet above water
height of tide = 2 feet = amount submerged in water
∴ length above water = (ladder before rising of the tide) - (amount submerged in water)
length above water = 8 - 2 = 6 feet
Therefore at the end of 6 hours, 6 feet of the ladder remained above water
Answer:
protection, support, and muscle attachment
Explanation:
hope this helps!
Maximum Thrift is a method / philosophy that considers the shortest tree (in number of transformation steps) the best hypothesis about the phylogenetic relationship of a given set of terminals. Transformation step is the cost attributed to the change of character state in a given branch of the tree, be it the change from one nucleotide base to another, amino acids, or the color of an animal's iris. The fewer transformations, the shorter the tree, and therefore the more parsimonious. In this method, what programs do is to investigate as many alternative topologies as possible (or all if their array has less than 25 terminals; see more details below) given their character array. Each will be measured, and the shorter (optimal) trees will be retained as the most parsimonious result.
The answer is C. I know that because of my previous knowledge and my class knowledge hope this helps :)<span />
