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3 years ago

Identify the minimum value of the function y = 2x2 + 4x Answer

Mathematics

2 answers:

iren2701 [21]3 years ago

4 0

Answer:

(-1, -2)

Step-by-step explanation:

Find the <em>x</em>-value the minimum occurs at. Use this value to find the minimum value. (Graph to find the highest point)

11111nata11111 [884]3 years ago

4 0

Answer:

y = - 2 is the minimum value

Step-by-step explanation:

The minimum value is the y- coordinate of the vertex.

The vertex lies on the axis of symmetry which is positioned at the midpoint of the zeros.

To find the zeros let y = 0, that is

2x² + 4x = 0 ← factor out 2x on the left side

2x(x + 2) = 0

Equate each factor to zero and solve for x

2x = 0 ⇒ x = 0

x + 2 = 0 ⇒ x = - 2

midpoint = $\frac{0-2}{2}$ = - 1 ← x- coordinate of vertex

Substitute x = - 1 into the equation for corresponding y- coordinate

y = 2(- 1)² + 4(- 1) = 2 - 4 = -2

vertex = (- 1, - 2)

Hence minimum value = - 2

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Answer:

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Answer:

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Step-by-step explanatio

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Ksenya-84 [330]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$