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Oxana [17]
2 years ago
10

(04.06 MC) In the graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded and labeled B, and the area

where f(x) and g(x) have shading in common is labeled AB. Graph of two intersecting lines. One line f of x is solid and goes through the points 0, 4, and 4, 0 and is shaded above the line. The other line g of x is solid, and goes through the points 0, negative 1 and 2, 5 and is shaded in below the line. The graph represents which system of inequalities? y ≤ −3x − 1 y ≤ −x − 4 y ≥ −3x + 1 y ≤ −x − 4 y ≤ 3x − 1 y ≤ −x + 4 y ≤ 3x − 1 y ≥ −x + 4

Mathematics
1 answer:
KonstantinChe [14]2 years ago
6 0

Answer:

(D)y ≤ 3x − 1 y ≥ −x + 4

Step-by-step explanation:

The line f(x) is solid and goes through the points (0, 4) and (4, 0) and is shaded above the line.

The line that satisfies the point (0,4) and (4,0) is y=-x+4

Since it is shaded above the line, we have the inequality sign  ≥.

Therefore, one of the lines is y≥-x+4

In the equation of the line y=3x-1

When x=0, y=3(0)-1=-1

When x=2, y=3(2)-1=5

Since it is shaded below the line, we have the inequality sign ≤ .

Therefore, the other line is y≤3x-1.

The correct option is D.

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Answer:

Step-by-step explanation:

a) The game tree for k = 5 has been drawn in the uploaded picture below where C stands for continuing and S stands for stopping:

b) Say we were to use backward induction we can clearly observe that stopping is optimal decision for each player in every round. Starting from last round, if player 1 stops he gets $3 otherwise zero if continues. Hence strategy S is optimal there.

Given this, player 2’s payoff to C is $3, while stopping yields $4, so second player will also chooses to stop. To which, player 1’s payoff in k = 3 from C is $1 and her payoff from S is $2, so she stops.

Given that, player 2 would stop in k = 2, which means that player 1 would stop also in k = 1.

The sub game perfect equilibrium is therefore the profile of strategies where both players always stop: (S, S, S) for player 1, and (S, S) for player 2.

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i hope this helps, cheers

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2 years ago
In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline
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Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

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And so, ∠BPE = 90° ⇒(1)

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2 years ago
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2 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
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Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
2 years ago
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