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lina2011 [118]
3 years ago
11

Frank owns a clothing store that sells shorts and graphic T-shirts. He sells the shorts for $14 each and the T-shirts for $6 eac

h. He is limited to the constraints shown by the set of inequalities below. Which of the points (s,t) will maximize Frank's revenue?
s≤ 18-0.5t
s≥ 10-t
s≤ 30-t
s≥0
t≥0
Mathematics
2 answers:
Dmitrij [34]3 years ago
6 0

Answer:

(18,0)

Step-by-step explanation:

Rasek [7]3 years ago
3 0

Answer:

Let s be the number of shorts and t be the number of T-shirts. s shorts cost $12s and t T-shirts cost $5t, then you have to find min and max value for the function f(s,t)=12s+5t.

The shaded domain (see image) is defined from the system of unequalities. The green lines are the graphs of function f(x,y) and it intersects domain in first point (0,5) (the minimum point) and in last point (20,0) (the maximum point). So,

.

Step-by-step explanation:

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Recycled​ CDs, Incorporated, offers a choice of 5 used CDs for ​$27​, with each additional CD costing ​$5. Write a cost function
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A function assigns the values. The cost of 6 CDs will be $32.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

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1 year ago
Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required
xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

                A = Pe^{rt}

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               6000 = 3000e^{0.085t}

               \frac{6000}{3000} = e^{0.085t}

               2 = e^{0.085t}

Take log_{e} of both sides

               log_{e}2 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln2 = 0.085t

               t = \frac{ln2}{0.085}

               t = \frac{0.693}{0.085}

               t = 8.15

(b) For the amount to double, A = 3 × P

               A = 3 × $3000

               A = $9000

               9000 = 3000e^{0.085t}

               \frac{9000}{3000} = e^{0.085t}

               3 = e^{0.085t}

Take log_{e} of both sides

               log_{e}3 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln3 = 0.085t

               t = \frac{ln3}{0.085}

               t = \frac{1.0986}{0.085}

               t = 12.92

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