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svetoff [14.1K]
3 years ago
10

(-5.5)(-4.87) Help help help help

Mathematics
1 answer:
kotegsom [21]3 years ago
6 0
(-5.5)(4.87) =26.785
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Mr. Martin is giving a math test next period. The test, which is worth 100 points, has 29 problems. Each problem is
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Answer:

x - the number of 5 points worth problems

y - the number of 2 points worth problems

The test has 29 problems.

Step-by-step explanation:

x - the number of 5 points worth problemsy - the number of 2 points worth problemsThe test has 29 problems.The test is worth 100 points.The system of equations:The solution:There are 14 problems worth 5 points and 15 problems worth 2 points.

x - the number of 5 points worth problems

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The bakery sells 8 doughnuts for every 3 bagels sold. If the bakery sells 9 bagels in a day, how many doughnuts are sold?
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Answer:

24. here is the explanation

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Step-by-step explanation:

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4 years ago
The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the o
Korvikt [17]

Answer:

a) k = 0.000124

b) According to these data, the Shroud of Turin has around 760 years.

Step-by-step explanation:

The amount of carbon-14 is modeled by the following equation:

C(t) = C_{0}e^{-kt}

In which C_{0} is the initial amount and k is the rate of decrease.

(a) Find the value of the constant k in the differential equation.

Half-life of 5595 years.

So C(5595) = 0.5C_{0}

C(t) = C_{0}e^{-kt}

0.5C_{0} = C_{0}e^{-5595k}

e^{-5595k} = 0.5

Applying ln to both sides

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b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?

This is t when C(t) = 0.91C_{0}

C(t) = C_{0}e^{-kt}

0.91C_{0} = C_{0}e^{-0.000124t}

e^{-0.000124t} = 0.91

Applying ln to both sides

-0.000124t = -0.094

t = 760.57

According to these data, the Shroud of Turin has around 760 years.

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4 years ago
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