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Gnesinka [82]
2 years ago
14

A video rental company offers a plan that includes a membership fee of $5 and charges $1 for every DVD borrowed. They also offer

a second plan, that costs $47 per month for unlimited DVD rentals. If a customer borrows enough DVDs in a month, the two plans cost the same amount. How many DVDs is that? What is that total cost of either plan?
Mathematics
2 answers:
DochEvi [55]2 years ago
8 0
That would be 42 dvds bc if each dvd is a dollar and there was first a 5 dollar fee, just subtract 47-5=42 and then 42divided by 1 which is 42$ so if you dont get 42 dvds per month, then go for the one with a 5 dollar fee
WINSTONCH [101]2 years ago
3 0

Answer:

for $5 plan if he rents 47 dvd it will pay 47% as one DVS cost 1 $ .

Step-by-step explanation:

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Answer:

1/250

Step-by-step explanation:

1000/4 simplifies to 1/250

All you had to do was simplify it

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Which of the following of x and y would make up the following expression represent a real number? (4+5i)(x+yi)
salantis [7]

Answer:

x = 4 and y = - 5

Step-by-step explanation:

note that the product of a complex number and it's conjugate is real

That is

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For (4 + 5i)(x + yi) to be real

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2 years ago
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Step-by-step explanation:

5 0
2 years ago
Make n the subject of the formula
kogti [31]

Answer:

\mathsf{ \frac{m + 21}{5} = n }

Step-by-step explanation:

To make "n" the subject of the formula, rearrange the formula so it begins with " n = "

To isolate the variable "n", you need to inverse the other terms on that side of the equation where "n" really is.

=> m = 5n - 21

  • <em>there's</em><em> </em><em>a</em><em> </em><em>"</em><em>-21</em><em>"</em><em> </em><em>next</em><em> </em><em>to</em><em> </em><em>"</em><em>5n</em><em>"</em><em>,</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>subtraction</em><em> </em><em>is</em><em> </em><em>addition</em><em>,</em><em> </em><em>so</em><em> </em><em>add</em><em> </em><em>21</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>.</em><em> </em>

=> m + 21 = 5n <u>-</u> <u>21</u> + <u>21</u>

=> m+ 21 = 5n

  • <em>There's</em><em> </em><em>also</em><em> </em><em>a</em><em> </em><em>5</em><em> </em><em>attached</em><em> </em><em>to</em><em> </em><em>n</em><em>,</em><em> </em><em>that</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>multiplication</em><em> </em><em>of</em><em> </em><em>n</em><em> </em><em>with</em><em> </em><em>5</em><em>,</em><em> </em><em>the</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>multiplication</em><em> </em><em>is</em><em> </em><em>division</em><em>,</em><em> </em><em>so</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em> </em><em>by</em><em> </em><em>5</em>

<em>\mathsf{ \frac{m + 21}{5} = n }</em>

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