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3 years ago

Is anyone willing to help me with this? I tried adding this and still no answer. Please help..

Mathematics

1 answer:

GaryK [48]3 years ago

3 0

Answer: C. (-4, -2)

Step-by-step explanation:

First, eliminate one of the variables and solve for the remaining variable:

2x - 5y = 2 → 3(2x - 5y = 2) → 6x - 15y = 6

3x + 2y = -16 → -2(3x + 2y = -16) → -6x - 4y = 32

-19y = 38

y = -2

Next, replace "y" with -2 into either of the original equations to solve for x:

2x - 5y = 2

2x - 5(-2) = 2

2x + 10 = 2

2x = -8

x = -4

x = -4, y = -2

Check:

Plug in the x- and y-values into the other original equation:

3x + 2y = -16

3(-4) + 2(-2) = -16

-12 + -4 = -16

-16 = -16 $\checkmark$

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3 years ago

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Suppose that you take 120 mg of an antibiotic every 4 hr. The half-life of the drug is 4 hr (the time it takes for half of the

vodomira [7]

Answer:

The steady state amount of antibiotic in the bloodstream when t --> ∞ is 240 mg.

Step-by-step explanation:

Let the amount of antibiotic in one's bloodstream be given as Aₙ (where n = the number of half lives since the start of usage)

Let's follow the time line of events.

At t = 0 hr, the drug is taken

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At t = 4 hrs, n = 1, the drug is taken again

A₁ = (0.5×A₀) + 120

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At t = 8 hrs, n = 2, the drug is taken again,

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At t = 12 hrs, n = 3, the drug is taken again

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At this point, it becomes evident that at t = 4n hrs, n = n i.e. n half lives later, the general formula for the amount of the antibiotic in the bloodstream is

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Hope this Helps!!!

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3 years ago

HELP ASAP!!!!!!!!

Hoochie [10]

Answer:

Step-by-step explanation:

The set {1,2,3,4,5,6} has a total of 6! permutations

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b. The next 120, including the 124th, would begin with '2'

c. Then of the 5! numbers beginning with 2, there are 4!=24 including the 124th number, which have the second digit =1

d. Of these 4! permutations beginning with 21, there are 3!=6 including the 124th permutation which have third digit 3

e. Among these 3! permutations beginning with 213, there are 2 numbers with the fourth digit =4 (121th & 122th), 2 with fourth digit 5 (numbers 123 & 124) and 2 with fourth digit 6 (numbers 125 and 126).

Lastly, of the 2! permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5 digit 6 (number 124).

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