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marysya [2.9K]
3 years ago
14

an online movie stream plan charges an annual fee of $45 plus $2.50 per movie watched.Another plan has no annual fee but charges

$3.75 per movie watched.For how many movies is the cost of the plans the same?
Mathematics
1 answer:
Anastasy [175]3 years ago
7 0
Ah, great question. Let me help you on this one.

We need to balance both equations for this to work out. This is needed because question asks us for an x variable that will make both equations equal to each other.

Let's write it out so it's easier for us to understand.

Let's not forget to multiply both sides by 100.

4500+(250x)=375x

Let's start by subtracting 45 from both sides.

4500+(250x)-4500=375x-4500

Let's simplify it.

250x=375x-4500

Now, we need to subtract 375 from both sides.

250x - 375=375x-4500+(-375)

Simplify it once again.

-125x=-4500

Divide both sides by -125

x=36

This plan will be the same cost after 36 movies.


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Find the length of the third side. If necessary, round to the nearest tenth.
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4.5

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3 years ago
Portland's population in 2007 was about 568 thousand, and had been growing by
slega [8]

Answer:

a) \ \ P_n=(1+r)P_{n-1}\\\\b)\ \ \ P_n=1.011^n(568000)\\\\c)\ \ 626,771\\\\d)\ \ 19.10yrs\ later \ (2027 \ February)

Step-by-step explanation:

a. Given that the population starts at 568000 and grows at 1.1% each years.

-The recursive formula for the population takes the form:

P_n=(1+r)P_{n-1}

Where:

P_n is the population at the nth year.

r is the rate of growth

P_{n-1} is the population a year before the nth year.

Hence, the recursive formula is given by P_n=(1+r)P_{n-1}

b. The explicit formula of a population growth takes the form:

P_n=(1+r)^nP_o

-Given that r=1.1% and the initial population is 568000

P_n=(1+r)^nP_o\\\\r=1.1\%=0.011\\\\P_n=(1+0.011)^nP_o, P_o=568000\\\\P_n=1.011^n(568000)

Hence, the explicit formula is P_n=1.011^n(568000)

c. The population in 2016 can be determined using the explicit formula.

-We substitute the growth rate and initial population as follows:

P_n=(1+r)^nP_o\\\\=1.011^n(568000)\\\\n=2016-2007=9\\\\\therefore P_{2016}=(1+0.011)^9\times 568000\\\\=626,770.7721\approx626,771

Hence, the population in 2016 will be approximately 626,771

d. Given that the population after n years will be 700000.

#We substitute this value in the explicit formula to solve for n then add it to the initial year, 2007;

P_n=(1+r)^nP_o\\\\700000=1.011^n(568000)\\\\1.011^n=\frac{700000}{568000}=\frac{700}{568}\\\\n=\frac{log \ (700/568)}{log\ 1.011}\\\\=19.10\ years\\\\\#Add \ 19.10yrs \ to \ 2007\\\\=2007+19.10yrs\\\\=2026.10\approx2027

Hence, the population will get to 700000 after 19.10 years or in February the year 2027

5 0
3 years ago
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