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Vilka [71]
3 years ago
10

Which of the following inequalities is best represented by this graph?

Mathematics
2 answers:
romanna [79]3 years ago
8 0

Answer: x-2y > 3

Step-by-step explanation:

By the given diagram,

The related line of the inequality is passes through the point (0,-1.5) and (3,0)

Thus, the equation of the related line,

y-(-1.5)=\frac{0-(-1.5)}{(3-0)} (x-0)

y+1.5=\frac{(0+1.5)}{3}x

y+1.5=\frac{1.5}{3}x

y+\frac{15}{10}=\frac{15}{30}x

y+\frac{3}{2}=\frac{1}{2}x

\frac{2y+3}{2}=\frac{1}{2}x

2y+3=x\implies 2y+3-x=0\implies -x+2y=-3\implies x-2y=3

Thus, the related line of the given inequality is x-2y=3

Hence, the possible inequalities are,

x-2y ≥ 3, x-2y ≤ 3, x-2y > 3 or x-2y < 3

By the given diagram,

The inequality does not contains the origin,

Thus, the possible inequalities are x-2y ≥ 3 or x-2y > 3.

Again,

The doted line shows that the inequality does not contain the equal sign.

Hence, the inequality is,

x-2y > 3

⇒ First option is correct.

blondinia [14]3 years ago
7 0
Based on the graph we see that the slope is 1/2

By simplifying each of these expressions, that leaves on options 1 and 2 that fit this.

Since the shaded area is going downward, the choice including less than is what's pictured.

Choice One is the correct answer
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A family built a house in a beach resort area. In 1990, the house was 430 ft from the water. With erosion, the house was 400 ft
neonofarm [45]

We develop an equation for the given situation by first writing the general equation for lines,

                                y = mx + b

Substituting to this given the values given above,

(1990)                      430 = b

(2000)                      400 = m(10) + 430

The value of m from the equation in 2000 is -3. Thus, the equation of that relates the variables is,

                              y = -3x + 430

8 0
3 years ago
A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per
HACTEHA [7]

Answer:

A) Revenue function = R(x) = (580x - 10x²)

Marginal Revenue function = (580 - 20x)

B) Fixed Cost = 900

Marginal Cost function = (300 + 50x)

C) Profit function = P(x) = (-35x² + 280x - 900)

D) The quantity that maximizes profit = 4

Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

8 0
3 years ago
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