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svetoff [14.1K]
3 years ago
7

Which graph represent bad a function of x

Mathematics
2 answers:
Dennis_Churaev [7]3 years ago
8 0

Answer:

idk

Step-by-step explanation:

idk....,..................

swat323 years ago
7 0

Answer: there is nothing to show us so where do we find this?

Step-by-step explanation:

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<h3>120 </h3>

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2 years ago
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A concession stand at an athletic event is trying to determine how much to sell cola and iced tea for in order to maximize reven
Cerrena [4.2K]

Solution :

Demand for cola : 100 – 34x + 5y

Demand for cola : 50 + 3x – 16y

Therefore, total revenue :

x(100 – 34x + 5y) + y(50 + 3x – 16y)

R(x,y)  = $100x-34x^2+5xy+50y+3xy-16y^2$

$R(x,y) = 100x-34x^2+8xy+50y-16y^2$

In order to maximize the revenue, set

$R_x=0, \ \ \ R_y=0$

$R_x=\frac{dR }{dx} = 100-68x+8y$

$R_x=0$

$68x-8y=100$  .............(i)

$R_y=\frac{dR }{dx} = 50-32x+8y$

$R_y=0$

$8x-32y=-50$  .............(ii)

Solving (i) and (ii),

4 x (i)    ⇒       272x - 32y = 400

     (ii)   ⇒   (-<u>)     8x - 32y = -50   </u>

                        264x        = 450

∴   $x=\frac{450}{264}=\frac{75}{44}$

     $y=\frac{175}{88}$

So, x ≈  $ 1.70      and    y = $ 1.99

    R(1.70, 1.99) = $ 134.94

Thus, 1.70 dollars per cola

          1.99 dollars per iced ted to maximize the revenue.

Maximum revenue = $ 134.94

4 0
3 years ago
Find the length of the third side. If necessary, round to the nearest tenth.
joja [24]

Answer: 5

Step-by-step explanation:

use formula a^2 + b^2 = c^2

a^2 + 12^2 = 13^2

a^2 + 144 = 169

       - 144   -144

a^2 = 25

sqrt   sqrt

a = sqrt (25)

a = 5

8 0
3 years ago
Nas funções f(x) = -3x+9; f(x) = 2x-4 e f(x) = 5x-5, caso construamos seus respectivos gráficos, informe respectivamente os pare
Likurg_2 [28]

1st option

{(3,0) e (0,9)}; {(2,0) e (0,-4)}; {(1,0) e (0,-5)}

see screenshot

sorry btw, no hablo espanol

8 0
2 years ago
Find the missing term of the following sequence. . . . –45, __, –12, . . .
julsineya [31]
<span>(-45 + (-12) / 2 = -28.5 
</span>So the answer is -28.5
3 0
3 years ago
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