Answer:
An atom consists of two basic parts: the nucleus and the electrons. The nucleus is the central core of an atom and is made up of protons and neutrons. Electrons are very light, negatively charged particles that surround the positively charged nucleus. Early models of the atom depicted the electrons circling the nucleus in fixed orbits, much like planets revolving around the sun.
Current theory suggests that electrons are housed in orbitals. This is a valence orbital or valence orbit
Explanation:
The outermost orbital shell of an atom is called its valence shell, and the electrons in the valence shell are valence electrons. Valence electrons are the highest energy electrons in an atom and are therefore the most reactive. this is where your valence orbit gets its name because it is the outermost shell.
Answer:
Log
Explanation:
The git log command enables you to display a list of all of the commits on your current branch
You use the Alignment sheet in the format cells dialog box to position data in a cell by centering it, for example. Generally along the middle branch is a line or centerline of the definite pipeline on a scale of whatever is indicated.
A string parameter and returns new string with all the letters of the alphabet that are not in the argument string. The letters in the returned string should be in alphabetical order. The implementation should uses a histogram from the histogram function
Explanation:
The below code is written in python :
alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]
test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]
def histogram(s):
d = dict()
for c in s:
if c not in d:
d[c] = 1
else:
d[c] += 1
return d
def has_duplicates(s):
for v in histogram(s).values():
if v > 1:
return True
return False
def test_dups_loop():
for s in test_dups:
print(s + ':', has_duplicates(s))
def missing_letters(s):
r = list('abcdefghijklmnopqrstuvwxyz')
s = s.lower()
for c in s.lower():
if c in r:
r.remove(c) # the first matching instance
return ''.join(r)
def test_miss_loop():
for s in test_miss:
print(s + ':', missing_letters(s))
def main():
test_dups_loop()
test_miss_loop()
if __name__ == '__main__':
main()
<em><u>Answer</u></em>
5 hours
<em><u>Explanation</u></em>
The two working together can finish a job in
Also, working alone, one machine would take one hour longer than the other to complete the same job.
Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.
Their combined rate in terms of x is
This should be equal to 20/9 hours.
Multiply through by;
Factor to get:
It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.
Therefore the slower finish in 5 hours.
