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kirill115 [55]
4 years ago
9

2. A lunch stand makes s.75 profit on each chef's salad and $1.20 profit on each Caesar salad. On a typical weekday, it sells be

tween 40 and 60 chef's salads and between 35 and 50 Caesar salads. The total number sold has never exceeded 100 salads. How many of each type should be prepared in order to maximize profit?
Could you make a graph as the answer?

Mathematics
1 answer:
kow [346]4 years ago
8 0

Let's let

<u>x = the number of chef salads, x>=0</u>

<u>y = the number of Caesar salads, y>=0</u>

The constrains are:

40 <= x <= 60

35 <= y <= 50

x + y <= 100

The objective function here is F(x, y) = 0.75x + 1.20y

The corner points are (40, 35),  (60, 35), (60, 40), (50, 50) and (40, 50).

F (40, 35) = 0.75*40 + 1.20*35 = $72

F (60, 35) = 0.70*60 + 1.20*35 = $84

F (60, 40) = 0.75*60 + 1.20*40 = $93

F (50, 50) = 0.75*50 + 1.20*50 = $97.50

F (40, 50) = 0.75*40 + 1.20*50 = $90

Thus, we conclude to maximize the profit 50 Chef and 50 Caesar salads should be prepared.

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Ken drew a pair of intersecting rays and marked the angle between them. . . Which of these statements best compares the pair of intersecting rays with the<span>angle</span> I'd say c because the common endpoint is the intersection. 
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For what value of p will the pair of linear equations have infinitely many solutions (p-3)x + 3y = p , px+ py = 12
sleet_krkn [62]

Answer: p = 6


Step-by-step explanation:

HI !


 px + 3y - ( p-3)=0


a₁ = p , b₁ = 3 , c₁ = - (p-3)


=====================


12x + py - p=0


a₂ = 12  , b₂ = p , c₂ = -p


since , the equations have infinite solutions , 


a₁/a₂ = b₁/b₂ = c₁/c₂


p/12 = 3/p =  p-3/p



--------------------------------------


p/12 = 3/p


cross multiply ,


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p = √36

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for the value of p = 6 , the equations will have infinitely many solutions





5 0
3 years ago
Can someone help me with this?
aivan3 [116]

Step-by-step explanation:

The formula for the volume of a sphere is V = 4/3 πr³.

So

Given

Volume (v) = 57ft³

v =  \frac{4}{3} \pi {r}^{3}

57 =  \frac{4}{3}  \times 3.14 \times  {r}^{3}

57 = \frac{12.56}{3} \times    {r}^{3}

57 = 4.19 \times  {r}^{3}

{r}^{3}  =  \frac{57}{4.19}

{r}^{3}  = 13.6

r =  \sqrt[3]{13.6}

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7 0
3 years ago
I have a photo up could someone answer two?
dedylja [7]
The answer to number 2 is g.15
3 0
3 years ago
Read 2 more answers
Find the difference between8/15 and 2/3 Show all calculations in your final answer.
kherson [118]

Answer:

  • \frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}          

Step-by-step explanation:

  • Let the value of a number 'a' be = 8/15
  • Let the value of a number 'b' be = 2/3

The difference between the two numbers can be calculated by subtracting the numbers

a-b=\frac{8}{15}-\left(\frac{-2}{3}\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

         =\frac{8}{15}-\frac{-2}{3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

          =\frac{8}{15}-\left(-\frac{2}{3}\right)

\mathrm{Apply\:rule}\:-\left(-a\right)=a

          =\frac{8}{15}+\frac{2}{3}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

          =\frac{8+10}{15}

          =\frac{18}{15}

\mathrm{Cancel\:the\:common\:factor:}\:3

           =\frac{6}{5}

Thus,

  • \frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}                    
3 0
3 years ago
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