Answer:
Explanation:
1)<u> Principal quantum number, n = 2</u>
- n is the principal quantum number and indicates the main energy level.
<u>2) Second quantum number, ℓ</u>
- The second quantum number, ℓ, is named, Azimuthal quantum number.
The possible values of ℓ are from 0 to n - 1.
Hence, since n = 2, there are two possible values for ℓ: 0, and 1.
This gives you two shapes for the orbitals: 0 corresponds to "s" orbitals, and 1 corresponds to "p" orbitals.
<u>3) Third quantum number, mℓ</u>
- The third quantum number, mℓ, is named magnetic quantum number.
The possible values for mℓ are from - ℓ to + ℓ.
Hence, the poosible values for mℓ when n = 2 are:
- for ℓ = 1, mℓ = -1, 0, or +1.
<u>4) Fourth quantum number, ms.</u>
- This is the spin number and it can be either +1/2 or -1/2.
Therfore the full set of possible states (different quantum number for a given atom) for n = 2 is:
- (2, 0, 0 +1/2)
- (2, 0, 0, -1/2)
- (2, 1, - 1, + 1/2)
- (2, 1, -1, -1/2)
- (2, 1, 0, +1/2)
- (2, 1, 0, -1/2)
- (2, 1, 1, +1/2)
- (2, 1, 1, -1/2)
That is a total of <u>8 different possible states</u>, which is the answer for the question.
Answer: It's a tie between f(x) and h(x). Both have the same max of y = 3
The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.
For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
turns into
h(x) = 2*1+1
h(x) = 2+1
h(x) = 3
showing that h(x) maxes out at y = 3 as well
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Note: g(x) has all of its y values smaller than 0, so there's no way it can have a max y value larger than y = 3. See the attached image to see what this graph would look like if you plotted the 7 points. A parabola seems to form. Note how point D = (-3, -2) is the highest point for g(x). So the max for g(x) is y = -2
Answer:
The correct answer is two designs are used 12 times each and other four designs are used for 13 times each.
Step-by-step explanation:
Width of the quilt Lu is making is 18 rows of 20 squares each.
Thus the perimeter of the quilt is 20 + 20 + 18 + 18 = 76.
We can also say there are 76 square as the boundary which are to be bordered with toy designs.
Each square can hold 1 design of toy and there are 6 different designs of toys.
Thus number of times each design is used
= 12
.
Thus there are two designs used 12 times each and other four designs are used for 13 times each.