Please help me solve this
1 answer:
9514 1404 393
Answer:
a) q = 15 at minimum average cost; q = 10 at minimum marginal cost
b) marginal cost = average cost = 225 at q = 15
Explanation:
The average cost is the total cost divided by the number of units.
A = C/q = 300 -10q +1/3q^2
The marginal cost is the derivative of the total cost.
M = 300 -20q +q^2
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a) The minimum average cost is where its derivative is zero.
A' = 0 = -10 +2/3q ⇒ q = 15
The minimum marginal cost is where its derivative is zero.
M' = -20 +2q ⇒ q = 10
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b) The average cost for 15 units produced is ...
A(15) = 300 -10(15) +1/3(15^2) = 225
The marginal cost at that same production level is ...
M(15) = 300 -20(15) +15^2 = 225
The average cost and marginal cost are equal at the minimum average cost.
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Answer:
The sequence of transformations will map figure K onto figure K′
is the first sequence <u>option (1)</u>
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Step-by-step explanation:
See the attached figure
as shown in the figure the K' is the image of K by reflection over x-axis
<u>But </u> We need to know which sequence of transformations will give the same result.
So, we will test the options by any point from K and its image from K'
i.e: we will test the options using the points (6,5) , (6,-5)
(6,5) ⇒ (6,-5)
<u>option (1):</u>
Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y)
(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (6,-5)
<u>option (2):</u>
Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y)
(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (-10,-5)
<u>option (3):</u>
Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y)
(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (2,-3)
<u>option (4):</u>
Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)
(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (-14,-3)
As shown: The sequence of transformations will map figure K onto figure K′
is the first sequence <u>option (1)</u>
