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VikaD [51]
3 years ago
6

The half-life of silicon-32 is 710 years. If 30 grams is present now, how much will be present in 300 years? Round answer to thr

ee decimal places. Using A(t)=A0e^(ln(0.5)/T)t
Mathematics
1 answer:
kupik [55]3 years ago
8 0

Answer: 55.668 grams

General equation of exponential decay/growth:

N = Noe^(kt)

50 = 100e^(710k)

.5 = e^(710k)

ln(.5) = 710k

ln(.5)/710 = k

Therefore, our equation is now:

N = 100e^((ln(.5)/710)t)

Now, we substitute t with 600:

N = 100e^(ln(.5)/710)600

N = 100e^(-.585758)

N = 100(.556684)

N = 55.668 grams

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Jill says that 12.6666666 is less than 12.63 explain Her error
Ymorist [56]
Look at the last number in the decimal.

12.66
12.63

Compare 6 and 3. 6 is larger so therefore 12.6666666 is larger. (You also need to compare the tenths place too, but these happen to have the same tenths place, .6.
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3 years ago
A bakery made 140 buns. A restaurant bought 80 of the buns. The remaining buns were put into bags of 4 buns each. How many bags
Sedbober [7]

Answer:

A bakery made 140 buns. A restaurant bought 80 of the buns. The remaining buns were put into bags of 4 buns each. How many bags of buns

are there?

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3 years ago
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a minute hand of a clock is 22 feet long.Find the distance traveled by the the tip of the minute hand in one hour.
Allushta [10]

Answer:

138.23 feet

Step-by-step explanation:

The minutes and seconds hands of a clock rotates in a circular manner no matter the shape of the clock. Therefore the distance covered by the minute hand of the clock is calculated by the circumference of a circle:

     C = 2πr

     C = 2 * \frac{22}{7} * 22

     C = 138.23 feet

7 0
3 years ago
A man started from his home and drives 30km to the south and then 86km to the north . How far is he from his home now ??
ladessa [460]

Answer:

56km North of his home.

Step-by-step explanation:

30km south - 86km north

= 56km north

6 0
3 years ago
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Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)
Allisa [31]

Answer:

Given that we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

From trigonometric identities, we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}

1 -  sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}

1 +  sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

3 0
3 years ago
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