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3 years ago

Will and olly share £80 in the ratio 3 : 2

Mathematics

1 answer:

Lelu [443]3 years ago

7 0

Answer:

Will's share = 48

Olly's share = 32

Step-by-step explanation:

Will's share : Olly's share = 3:2

Will's share = 3x

Olly's share = 2x

3x + 2x = £ 80 {Combine like terms}

5x = 80 {Divide both sides by 5}

x = 80/5

x = 16

Will's share = 3x = 3*16 = 48

Olly's share = 2x = 2*16 = 32

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2 years ago

Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.

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Answer :

The amount after 1000 years will be, 5.19 grams.

The amount after 10000 years will be, 0.105 grams.

Step-by-step explanation :

Half-life = 1599 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{1599\text{ years}}$

$k=4.33\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount after 1000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 1000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$1000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=5.19g$

Thus, the amount after 1000 years will be, 5.19 grams.

Now we have to calculate the amount after 10000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 10000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$10000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=0.105g$

Thus, the amount after 10000 years will be, 0.105 grams.

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3 years ago