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Aleksandr-060686 [28]
3 years ago
8

The table below represents a linear function f(x) and the equation represents a function g(x):

Mathematics
2 answers:
Mademuasel [1]3 years ago
5 0

Answer:

Part A: The slope of g(x) is greater than the slope of f(x).

Part B: Function g(x) has greater y-intercept.

Step-by-step explanation:

The function f(x) passes through the points (-1,-3), (0,0) and (1,3). It means the y-intercept of the function is 0.

If a line passes through two points, then the slope of the function is

m=\frac{y_2-y_1}{x_2-x_1}

Let as consider any two points of the function, (-1,-3) and (0,0).

m=\frac{0+3}{0+1}=3

The point slope form of a linear function is

m=mx+b                .... (1)

Where, m is slope and b is y-intercept.

The slope of f(x) is 3 and y-intercept is 0, therefore the function f(x) is

f(x)=3x

The given function is

g(x)=7x+2            .... (2)

From (1) and (2), we get

m=7,b=2

The slope of g(x) is 3 and y-intercept is 2.

Part A: Since the slope of f(x) is 3 and the slope of g(x) is 7, therefore the slope of g(x) is greater than the slope of f(x).

Part B: Since the y-intercept of f(x) is 0 and y-intercept of g(x) is 2, therefore the function g(x) has greater y-intercept.

Eddi Din [679]3 years ago
3 0

A: Looking at the last two rows of the table of f(x), the slope is simply (3-0)/(1-0) = 3, while the slope of g(x) is the coefficient of the variable x, which is 7.


B. y-int is the y value when x = 0. f(x) has a y-int of 0, which g(x) is 2. The latter has a greater y-int.

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Blizzard [7]

Answer:

(a) C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2

    p(x)=320-7.7p

    R(x)=(320-7.7p)p=320p-7.7p^2

(b) x=82 \text{planes}

(c) p=\$30.91 M\;\; \text{per plane}

(d) maximum profit =\$ 15.90M

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

m(x)=20x-5x^{\frac{3}{4}}+0.01x^2

(a)  Find the cost, demand (or price), and revenue functions.

   C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2

   p(x)=320-7.7p

   R(x)=(320-7.7p)p=320p-7.7p^2

(b)  Find the production level that maximizes profit.

    f=R(x)-C(x)

 \Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)

\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx

     x=320-7.7p

     p=\frac{320-x}{7.7}

    \frac{dp}{dx} = \frac{-1}{7.7}

\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0

    \Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0

   \Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0

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(c)  Find the associated selling price of the aircraft that maximizes profit.

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(d)  Find the maximum profit.

Manufacturing cost of one plane is:

m(1)=20-5+0.01

         =\$15.01 M

maximum profit =\$(30.91-15.01)M

                           =\$15.90M

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Alecsey [184]

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