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Sever21 [200]
3 years ago
8

Laura rents a movie for a flat fee of $2.00 plus an additional $0.50 for each night she keeps the movie. Choose the cost functio

n that represents this scenario if x equals the number of nights Laura has the movie.
a) c(x) = 2.00x + 0.50
b)c(x) = 2.00 + 0.50x
c) c(x) = 2.50x
d)c(x) = (2.00 + 0.5
Mathematics
2 answers:
daser333 [38]3 years ago
8 0

Answer:

The cost function that represents this scenario is c(x) = 2 + 0.50x .

Option (b) is correct .

Step-by-step explanation:

As given

Laura rents a movie for a flat fee of $2.00 plus an additional $0.50 for each night she keeps the movie.

if x equals the number of nights Laura has the movie.


Than the cost function that represents this scenario .

c(x) =  Flat fee + Cost for  x equals the number of nights Laura has the movie.


c(x) = 2 + x × 0.50

c(x) = 2 + 0.50x

Therefore the cost function that represents this scenario is c (x) = 2 + 0.50x .

Option (b) is correct .

Semmy [17]3 years ago
8 0
B)c(x) = 2.00 + 0.50x

Remember that since 2.00 is the flat fee, so there is no variable. After that, though, there is the fee of .50 which is charged every day you have a movie.

say you had the movie rented out for five days
c(5)=2.00+.50(5)
=2.00+2.5
c(5)=4.5

Hope this helped!

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The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
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Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

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WITCHER [35]

f(n) = n² + 20

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