A movie rental store charges a membership fee of $10.00 plus $1.50 per movie rented. What is a possible domain that shows the cost to a member of renting movies?A. The domain is all integers.

B. The domain is all even integers.

C. The domain is all integers 0 or greater.

D. The domain is all integers 10 or greater.

Answers

Answer 1
Answer: The domain is all integers 10 or greater. This is because the working equation for the statement is movie renting = 10 + 1.5 x(movie renting). With that equation there is a minimum of 10.

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An astronaut who weighs 126lb on Earth weighs only21lb on the moon. How much would a person who weights 31lb on the moon weigh on Earth?
In an expression(8/9) squared (-81)+3/5dividedby-9/10
If y varies directly with x and x=4 when, y=20, find k, the constant of variation.

Present age of Ram and Ravi are in the ratio4:3.The ratio of their ages will be 5:4 after
5 years. Find Ravi's present age​

Answers

Answer:

15 years old

Step-by-step explanation:

Given the ratio of their ages = 4 : 3 = 4x : 3x ( x is a multiplier )

Then in 5 years the ratio of their ages is 4x + 5 : 3x + 5 = 5 : 4, that is

(4x+5)/(3x+5) = (5)/(4) ( cross- multiply )

4(4x + 5) = 5(3x + 5) ← distribute both sides

16x + 20 = 15x + 25 ( subtract 15x from both sides )

x + 20 = 25 ( subtract 20 from both sides )

x = 5

Thus

Ravi's age is 3x = 3 × 5 = 15

Answer:

Step-by-step explanation:

Ravi présent âge 15 because présent would be 20:15 and then it would turn into 25:20

Convert 54 3/4 into decimal notation.

Answers

Answer: 54 and 3/4 can be written as 54.75 in decimal notation.

Write the most precise name for the space figure with the given properties. A lateral surface and two circular bases?

Answers

The figure would be a cylinder, which is two circular bases connected by a lateral side.
That would be a cylinder. c;

How do i solve divide 5/8 by 50%

Answers

50% is (50)/(100)
We have two ways to do this, pick your favor:
either by fractions:
(5)/(8) ÷ (1)/(2)
((50)/(100) = (1)/(2)
Multiply by the reciprocal of (1)/(2) which is 2(flip it)

(5)/(8) × 2 = (5)/(4) = 1.25

Or change them to decimals:
(5)/(8) = 0.625
(50)/(100) = 0.5

0.625 ÷ 0.5

0.5 I 0.625
multiply by 10 to make it easier to divide(it still gives us the same answer)
0.5 × 10 = 5
0.625 × 10 = 6.25
     1.25
5 I 6.25
    -5
     12
    -10
       25
      -25
         0

0.625 ÷ 0.5 = 1.25

Find the greatest common factor of 14, 20, and 25.

Answers

14=2                       20=2                            25=5
 7=7                        10=2                              5=5  
 1=1                         5=5                               1=1        
so the one for 14     1=1                                so the answer for 25
would be 2*7           so the answer for 20      would be 5*5
                                would be 2*2*5 or 4*5

the reason that i did not put the one into the math problems is because anything times one is its self so i hope that this helps 

Answer: it 1

like pls

Step-by-step explanation:

Ali and Kiana buried a treasure together on their school's field. The actual field is 400 feet wide. Ali made an 8-inch-wide map to record its location. Kiana made her map using a scale of 1 in. To 20 ft. On Kiana's map, the treasure is 2 inches from the south edge of the field. How far is the treasure from the south edge on Ali's map?

Answers

Answer:

The treasure is 0.8 inches from the south edge on Ali's map.

Step-by-step explanation:

Scaling factor, f = (Original length)/(Scalet length)...(i)

Let f_1 and f_2 be the scaling factors used by Ali and Kiana respectively.

Given that the field is 400 feet= 400x12 inches wide and Ali made an 8-inch-wide map to record its location.

So, f_1 = (400*12)/8=600...(ii)

Kiana made her map using a scale of 1-inch to 20 feet=20x12 inches.

So, f_2=(20*12)/1=240...(iii)

As on Kiana's map, the treasure is 2 inches from the south edge of the field,

so, from equations (i), and (ii), the original length of the treasure for the south edge of the field

=2* f_2

=2x240

=480 inches

Now, again from the equation (i) and (ii), the scaled length of the treasure on Ali's map

= 480/f_1

=480/600

=0.8 inches

Hence, the treasure is 0.8 inches from the south edge on Ali's map.