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QveST [7]
2 years ago
5

The profitability, P, of a popular restaurant franchise can be modeled by the function P (t) = t4 – 10t3 + 28t2 – 24t, where t i

s the number of months since the restaurant opened. How many months after the franchise opens will it begin to show a profit?
A. 0 months
B. 2 months
C. 4 months
D. 6 months
Mathematics
1 answer:
vekshin12 years ago
8 0

In this question, the profit of the restaurant after t months is given by a polynomial function. To find when it begins to show a profit, we find the numerical values of the function for t, and it shows a profit when t > 0

Profit after t months:

P(t) = t^4 - 10t^3 + 28t^2 - 24t

0 months:

This is P(0). So

P(0) = 0^4 - 10*0^3 + 28*0^2 - 24*0 = 0

1 month:

This is P(1). So

P(1) = 1^4 - 10*1^3 + 28*1^2 - 24*1 = -5

2 months:

This is P(2). So

P(2) = 2^4 - 10*2^3 + 28*2^2 - 24*2 = 0

3 months:

This is P(3). So

P(3) = 3^4 - 10*3^3 + 28*3^2 - 24*3 = -9

4 months:

This is P(4). So

P(4) = 4^4 - 10*4^3 + 28*4^2 - 24*4 = -32

5 months:

This is P(5). So

P(5) = 5^4 - 10*5^3 + 28*5^2 - 24*5 = -45

6 months:

This is P(6). So

P(6) = 6^4 - 10*6^3 + 28*6^2 - 24*6 = 0

7 months:

This is `P(7). So

P(7) = 7^4 - 10*7^3 + 28*7^2 - 24*7 = 175

After 7 months it shows profit, so it starts showing profit on the 6th month, and thus, the correct answer is given by option D.

For another example of a function involving numeric value, you can check brainly.com/question/24231879

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The graph shows the quantity of each piece in a toy building kit. One piece is missing. What is the probability that the missing
Ivahew [28]

Answer:

2/35

Step-by-step explanation:

The only way we can solve this problem is to assume that the given complement of pieces is of a full kit before one went missing. The total number of pieces in the kit is ...

... 9 + 8 + 12 + 6 = 35

Of those, 2 are red with four holes. Thus the probability that a randomly chosen piece is red with 4 holes is 2 out of 35, or 2/35.

____

If this is the kit contents after the piece went missing, we have no way of knowing the color or hole count of the missing piece. If the kit is supposed to contain 13 blue pieces, for example, then the probability is zero that the missing piece is red.

8 0
2 years ago
f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0
2 years ago
A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the fun
IrinaVladis [17]
The answer is: 15.75cm3
7 0
2 years ago
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a person on a tour has rupees 12000 for his daily expenses. In order to extend his journey for 2 more days he had to cut down hi
myrzilka [38]

Answer:

Duration of the tour he planned first is 8 days.

Step-by-step explanation:

Given that a person has 12000 rupees for his daily expenses.

Let x be the number of days.

Then daily expenses per day =\frac{12000}{x}

Given that number of day is increased by 2 more days, that is number of days is x+2.

New daily expense per day \frac{12000}{x+2}

Given that this new daily expenses are 300 less than original.

That is \frac{12000}{x}-\frac{12000}{x+2}=300

              \frac{12000(x+2)-12000x}{x(x+2)}=300

                  \frac{24000}{x(x+2)}=300

                  x(x+2)=\frac{24000}{300}

                    x^{2}+2x-80=0

                    (x-8)(x+10)=0

                     x=8 or -10.

Since number of days cannot be negative, duration of the tour planned=8.

7 0
2 years ago
The profit earned by a sporting goods outlet is modeled in the graph below, where x is the number of years since the outlet open
LenaWriter [7]

The domain would be x ≥ 0.

This is because the outlet cannot have profit before it was open. Therefore, the growth must be from year 0 to present. If they give a year as starting, you can have an upper limit too, but there is not enough information here to determine that information.

4 0
2 years ago
Read 2 more answers
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