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Alina
11 days ago
12

The graph of f(x) = |x – h| + k contains the points (–6, –2) and (0, –2). The graph has a vertex at (h, –5). Describe how to fin

d the value of h. Then, explain how this value translates the graph of the parent function.
Mathematics
2 answers:
lawyer [12.5K]11 days ago
5 0
<span> The absolute value function exhibits symmetry. Given that the coordinates (–6, –2) and (0, –2) produce the same output, the points are equidistant from the line of symmetry. The value of –3 exists between –6 and 0. Therefore, the x-coordinate of the vertex must be –3, which is the value of </span>h<span>. This indicates that the graph of the parent function shifts 3 units to the left.</span>
AnnZ [12.3K]11 days ago
3 0
The function representing absolute value displays symmetry, with its vertex aligned on the line of symmetry. The coordinates (–6, –2) and (0, –2) yield equivalent outputs, indicating they are equidistant from the line of symmetry. The midpoint between –6 and 0 is –3. Consequently, the x-coordinate of the vertex must be –3, which corresponds to h. This results in a leftward shift of the graph of the parent function by 3 units.
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Arrange them starting from the smallest to the largest.
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Answer:

4.0921 reflects the logarithm of the equilibrium constant.

Step-by-step explanation:

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Iron has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the anode.

E^{o}_{cell}=Reduction potential of cathode - Reduction potential of anode

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2Ag^+(aq) + 2e^-\rightarrow 2Ag(s); E° = 0.80 V

Net reaction: Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)

n = 2

To determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

\Delta G^o=-nfE^o_{cell}

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\Delta G^o=-RT\ln K_{eq}

Aligning these two equations yields:

nfE^o_{cell}=RT\ln K_{eq}

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n = electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 1.21 V

R = gas constant = 8.314 J/K.mol

T = reaction temperature = 25^oC=[273+25]=298K

Substituting values into the equation, we arrive at:

2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}

\ln K_{eq}=9.3478

\log K_{eq}=\frac{9.3478}{2.303}=4.0921

4.0921 represents the logarithm of the equilibrium constant.

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