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Anastaziya [24]
3 years ago
8

The total number of f orbitals in an f subshell is

Mathematics
1 answer:
Vladimir79 [104]3 years ago
5 0
There are seven f<span> orbitals in an </span>f<span> subshell</span>
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Answer:

the answer is c - (6,3)

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Through: (1,0), slope=1
vitfil [10]
Standard form:
Y - 0= 1(x - 1)
Y - 0= x - 1
-x+y= -1
- (-x+y= -1)
Answer:
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Slope-Intercept form:
Y - 0= 1(x-1)
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4 0
3 years ago
The congruence theorem that can be used to prove ALON
Arisa [49]

Answer:

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN ⇒ 1st answer

Step-by-step explanation:

Let us revise the cases of congruence

  • SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
  • SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
  • ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
  • AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
  • HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ  

In triangles LON and LMN

∵ LO ≅ LM ⇒ given

∵ NO ≅ NM ⇒ given

∵ LN is a common side in the two triangles

- That means the 3 sides of Δ LON are congruent to the 3 sides

   of Δ LMN

∴ Δ LON ≅ LMN ⇒ by using SSS theorem of congruence

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN

3 0
3 years ago
The height of a punted football can be modeled with the quadratic function 0.01x^2 + 1.18x+2.
Fantom [35]

The function of the path of the punted ball is a quadratic function which

follows the path of a parabola.

The correct responses are;

Part A: The coordinates of the vertex is \underline {(59, \, 36.81)}

Part B: The maximum height of the punt is <u>36.81 ft.</u>

Part C: The defensive player must reach up to <u>7.65 feet</u> to block the punt.

Part D: The distance down the field the ball will go without being blocked is approximately <u>119.67 ft.</u>

<u />

Reasons:

The function for the height of the punted ball is; h = -0.01·x² + 1.18·x + 2

Assumption; The distances are feet.

Part A: By completing the square, we have;

f(x) = -0.01·x² + 1.18·x + 2

100·f(x) = -x² + 118·x  + 200

-100·f(x) = x² - 118·x  - 200

x² - 118·x + (118/2)²= 200 + (118/2)²

(x - 59)² = 200 + (59)² = 3681

(x - 59)² - 3681

At the vertex, -3281 = -100·f(x)

∴ f(x) at the vertex = -3681/-100 = 36.81

\mathrm{\underline{Coordinate \ of \ the \ vertex = (59, \, 36.81)}}

Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.

Part C: When <em>x</em> = 5, we have;

h = -0.01·x² + 1.18·x + 2

h = -0.01 × 5² + 1.18 × 5 + 2 = 7.65

The defensive player must reach up to 7.65 feet to block the punt

Part D: The distance the ball will go before it hits the ground is given by

the function, for the height as follows;

h = -0.01·x² + 1.18·x + 2 = 0

From the completing the square method, above, we get;

-0.01·x² + 1.18·x + 2 = 0

x² - 118·x  - 200 = 0

x² - 118·x + (118/2)²= 200 + (118/2)²

x² - 118·x + (59)²= 200 + (59)² = 3681

(x - 59)² = 3681

x - 59 = ±√3681

x = 59 ± √3681

x = 59 + √3681 ≈ 119.67

The distance down the field the ball will go without being blocked, x ≈ <u>119.67 ft.</u>

<u />

Learn more here:

brainly.com/question/24136952

5 0
2 years ago
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