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Lisa [10]
3 years ago
8

A bakery gave a doughnut to every fifth customer and a cupcake to every twelfth customer. In one day, 8 doughnuts and 3 cupcakes

were given to customers. Which could be the number of customers the bakery had that day?
Mathematics
2 answers:
cricket20 [7]3 years ago
6 0

The bakery had seventy - six(6) customers that day

steposvetlana [31]3 years ago
5 0

Answer:

76

Step-by-step explanation:

Step 1: Identify given information and create 2 equations.

c = # of customers g = given items

1/5c = g (Equation for doughnuts)

1/12c = g (Equation for cupcakes)

Step 2: Substitute in for given items

1/5c = 8

1/5c = 3

Step 3: Use PEMDAS to find c in each equation.

1/5c = 8

*5 *5

c = 40

1/12c = 3

*12 *12

c = 36

Step 4: add customers from both equations

40 + 36 = 76

The bakery had 76 customers that day.

(note I am not a professional and please know that my advice may be incorrect at times)

Hope this helps =)

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Step-by-step explanation:

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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


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Step-by-step explanation:

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Please help me on #37. I know for A you would use the quadratic equation but I got the wrong answer.
ozzi
Δ = (8i)^2 - 4*(-25) => Δ = -36 +100 => Δ = 64 => x1 =(-8i + 8)/2 => x1 = -4i +4;
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3 years ago
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