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7nadin3 [17]
3 years ago
13

Can someone pleaseeee help and if you’re correct I’ll give u brainliest

Mathematics
1 answer:
liq [111]3 years ago
6 0

Answer:

The answer is b

Step-by-step explanation:

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Lex is at the mall, which is 8 miles from his house. Lex walks home at a constant rate of 2 miles an hour.
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Y = -2x + 8 <=== this would be ur equation where x is the number of hrs and y is the total distance (in miles) from home
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Use the multiplier method to decrease £27 by 8%. You must show your working.
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Answer:

decrease by 2.16

Step-by-step explanation:

Convert the problem to an equation using the percentage formula: P% * X = Y.

P is 10%, X is 150, so the equation is 10% * 150 = Y.

Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.

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The antibiotic clarithromycin is eliminated from the body according to the formula A(t) = 500e−0.1386t, where A is the amount re
PolarNik [594]

Answer:

Time(t) = 11.61 hours (Rounded to two decimal place)

Step-by-step explanation:

Given: The antibiotic  clarithromycin is eliminated from the body according to the formula:

A(t) = 500e^{-0.1386t}                 ......[1]

where;

A - Amount remaining in the body(in milligram)

t - time in hours after the drug reaches peak concentration.

Given: Amount of drug in the body is reduced to 100 milligrams.

then,

Substitute the value of A = 100 milligrams in [1] we get;

100= 500e^{-0.1386t}

Divide both sides by 500 we get;

\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}

Simplify:

\frac{1}{5} = e^{-0.1386t}

Taking logarithm both sides with base e, then we have;

\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})

\log_e (\frac{1}{5})=-0.1386t         [ Using \log_e e^a =a ]

or

\log_e (0.2)=-0.1386t

-1.6094379124341 = -0.1386t

 [using value of \log_e (0.2) = -1.6094379124341 ]

then;

t = \frac{-1.6094379124341}{-0.1386}

Simplify:

t ≈11.61 hours.

Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams


6 0
3 years ago
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