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2 years ago

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PLZ HELP ME GUYS WITH THIS EQUATION: Manuel will choose between two restaurants to purchase pizzas for his party. The first rest

aurant charges a delivery fee of $2 for the entire purchase and $10 per pizza. The second restaurant charges a delivery fee of $5 for the entire purchase and $9 per pizza. Let x be the number of pizzas purchased.

1 answer:

pishuonlain [190]2 years ago

3 0

Answer:

x = 3 pizzas

Step-by-step explanation:

Not sure what the question is asking, but I am assuming that the question is asking how many pizzas needs to be ordered for the restaurant ordered to equal each other.
Set up equation:
$10x + 2 = 9x + 5$
Simplify the equation:

$x = 3$

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Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

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Answer:

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Answer:

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2 years ago

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Answer:

Option C.

Step-by-step explanation:

Given information:

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