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3 years ago

Help im not sure if i am doign this right or not

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

3 0

Answer:

(a) X = {Bilk, Jas, Piaf, Sydow, Wil}

(b) Y = {Jas, Sydow, Tixo}

Hope that helps.

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Answer:

d because it does not looks somlar

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2 years ago

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Plz help 15 + (-8) =

Effectus [21]

Answer:

Step-by-step explanation:

15 + (-8)

The answer would be 7 because you are only taking 15 plus negative 8 which would technically be the same as taking 15 minus 8, which would be 7.

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3 years ago

Use the distributive property to simplify this expression: negative 2 (3 x + x minus 5). 1. Using order of operations, combine l

sashaice [31]

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8x-10

Step-by-step explanation:

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3 years ago

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Estimate the product of 26.4×2.7

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3 years ago

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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

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

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago