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vivado [14]
3 years ago
12

Ray needs to mail his Christmas cards and packages and wants to keep the mailing costs to no more than $400 . The number of card

s is at least 20 more than twice the number of packages . The cost of mailing a card is $ 1 and a package cost is $ 5. a . Write a system of inequalities to model the problem . b Graph the system . Can he mail 120 cards and 50 packages ? d . Can he mail 220 cards and 100 packages ?
Mathematics
2 answers:
jeka943 years ago
5 0
The anwser for your question is 120
Elina [12.6K]3 years ago
5 0

Answer:

yes he can mail 120 cards and 50 package

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3 years ago
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The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
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Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

6 0
3 years ago
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D is correct.This is because these lines will never intersect
<span>Their slopes are equal
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3 years ago
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Me ayudan por favor ❤️
Ghella [55]

Answer:

Step-by-step explanation:

ABC = Right Angle

EFG = Acute Angle

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The answer is        A .2d/T
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