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Mekhanik [1.2K]
3 years ago
5

The half-life is a substance is 375 years. If 70 grams is present now, how much will be present in 500 years?

Mathematics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

27.76 grams will be present in 500 years

Step-by-step explanation:

The given formula is A=A_{o}e^{kt} , where A is the value of the substance in t years, and A_{o} is the initial value

∵ The half-life is a substance is 375 years

- Substitute A by \frac{1}{2}A_{o} and t by 375 to find the value of k

∴ \frac{1}{2}A_{o}=A_{o}e^{375k}

- Divide both sides by A_{o}

∴ \frac{1}{2}=e^{375k}

- Insert ㏑ in both sides

∴ ㏑( \frac{1}{2} ) = ㏑ ( e^{375k} )

- Remember ㏑ ( e^{n} ) = n

∵ ㏑ ( e^{375k} ) = 375 k

∴ ㏑( \frac{1}{2} ) = 375 k

- Divide both sides by 375

∴ k ≈ -0.00185

∴  A=A_{o}e^{-0.00185t}

∵ 70 grams is present now

- That means the initial value is 70 grams

∴ A_{o} = 70

∵ The time is 500 years

∴ t = 500

- Substitute the values of A_{o} and t in the formula

∵ A=70e^{-0.00185(500)}

∴ A = 27.76

∴ 27.76 grams will be present in 500 years

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Step-by-step explanation:

The cost of an adult ticket is £6 more than that of a child ticket, so will be shown by c+6.

Now, we are told that the cost of four child tickets and two adult tickets is £40.50, so we can put this in an equation and solve for c:

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A homeowner has two smoke detector alarms installed, one in the dining room (adjacent to the kitchen) and one in an upstairs bed
cestrela7 [59]

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Step-by-step explanation:

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Alexis, Becca, and Cindy do volunteer work at Adopt-a-Pet Animal Shelter. They worked a total of 285 hours at the shelter last s
swat32

Alexis worked for 59 hours, Becca worked for 74 hours and Cindy worked for 152 hours.

Step-by-step explanation:

Given,

Total hours worked by three of them = 285 hours

Let,

x represent the hours of Alexis.

y represent the hours of Becca.

z represent the hours of Cindy.

According to given statement;

x+y+z=285    Eqn 1

Alexis worked 15 hours less than Becca.

x = y-15         Eqn 2

z = 2y+4        Eqn 3

Putting value of x and z from Eqn 2 and 3 in Eqn 1

(y-15)+y+(2y+4)=285\\y-15+y+2y+4=285\\4y-11=285\\4y=285+11\\4y=296

Dividing both sides by 4

\frac{4y}{4}=\frac{296}{4}\\y=74

Putting y=74 in Eqn 2

x=74-15\\x=59

Putting y=74 in Eqn 3

z=2(74)+4\\z=148+4\\z=152

Alexis worked for 59 hours, Becca worked for 74 hours and Cindy worked for 152 hours.

Keywords: linear equation, division

Learn more about division at:

  • brainly.com/question/9510228
  • brainly.com/question/9443926

#LearnwithBrainly

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Step-by-step explanation:

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