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Luda [366]
3 years ago
7

Find the rule for each table in the pic please

Mathematics
2 answers:
zimovet [89]3 years ago
4 0
(1,3)(2,7)
slope = (7 - 3) / (2 - 1) = 4

y = mx + b
slope(m) = 4
use any of ur points (1,3)...x = 1 and y = 3
now we sub and find b, the y int
3 = 4(1) + b
3 = 4 + b
3 - 4 = b
-1 = b

so ur rule is : y = 4x - 1 or f(x) = 4x - 1
=============================
(2,8)(3,11)
slope = (11 - 8) / (3 - 2) = 3

y = mx + b
slope(m) = 3
use any of ur points....(2,8)...x = 2 and y = 8
now sub and find b, ur y int
8 = 3(2) + b
8 = 6 + b
8 - 6 = b
2 = b

so ur rule is : y = 3x + 2 or f(x) = 3x + 2

coldgirl [10]3 years ago
3 0
(left table) the first one is add 1(on left) and add 4 (on right)
(right table) then the second on is 
add 1(on left) and add 3 (on right)
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An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co
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

   \Rightarrow x=82 \text{planes}

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

  p=\frac{320-82}{7.7}

\Rightarrow p=\$30.91 M\;\; \text{per plane}

(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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