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ASHA 777 [7]
3 years ago
6

A farmer is comparing the number of eggs laid by two hens. The first hen laid 7 eggs per week for x weeks. The second hen laid 8

eggs per week for y weeks. The total number of eggs laid by the first hen was at least the total number of eggs laid by the second hen. Write an inequality showing this relationship
Mathematics
1 answer:
nikitadnepr [17]3 years ago
7 0

Answer:

7x ≥ 8y

Step-by-step explanation:

The first hen laid 7 eggs per week for x weeks.  7x

The second hen laid 8 eggs per week for y weeks. 8y

The total number of eggs laid by the first hen was at least the total number of eggs laid by the second hen.

7x ≥ 8y

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A side of the triangle below has been extended to form an exterior angle of 129°. Find the value of x.
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A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place
Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

P(t) = P(0)(1-r)^{t}

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

P(1) = 0.97P(0)

Then

P(t) = P(0)(1-r)^{t}

0.97P(0) = P(0)(1-r)^{0}

1 - r = 0.97

So

P(t) = P(0)(0.97t)^{t}

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

P(t) = P(0)(0.97t)^{t}

0.5P(0) = P(0)(0.97t)^{t}

(0.97)^{t} = 0.5

\log{(0.97)^{t}} = \log{0.5}

t\log{0.97} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.97}}

t = 22.76

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

P(t) = P(0)(0.97t)^{t}

0.85P(0) = P(0)(0.97t)^{t}

(0.97)^{t} = 0.85

\log{(0.97)^{t}} = \log{0.85}

t\log{0.97} = \log{0.85}

t = \frac{\log{0.85}}{\log{0.97}}

t = 5.34

5.34 years for the sample to decay to 85% of its original amount

8 0
4 years ago
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