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2 years ago

Evaluate 2^-4/4^0 Write your answer as a fraction in simplest form

Mathematics

1 answer:

Luden [163]2 years ago

5 0

Answer:

1/16 or 0.0625

Step-by-step explanation:

Im not sure but this may be the answer

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Sam invest $8333 into a savings account with a compound interest of 2.7% for 10 years what is Sams balance after 10 years

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Answer: 10,582.91

Step-by-step explanation: 2.7%*10=27% 8,333*127%= $10,582.91

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Step-by-step explanation:

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What is 3/5 × 4/9 as a fraction.<br><br> Plz help :(

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3 years ago

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200 kg of brass is melted down and cast into ornamental frogs, each weight 3/20kg. How many frogs are made?

Savatey [412]

Answer:

1,333 frogs

Step-by-step explanation:

okay so we know that mass remains conserved no matter where you are.

same here ^^

if the total mass of brass is 200 kg

the total mass of the new frogs formed from it when put together will be the same :)

and if there were n such frogs formed

we have,

total mass = n × mass of each frog

200 = n × 3/ 20

n = 4000/ 3

so it becomes something like 1,333.33

that is nearly 1,333 frogs can be made out of 200 kg of brass each weighing 3/ 20kg (the last ones a bit less to make upto 200kg)

5 0

3 years ago

Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour

UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

$C(t) = C(0)e^{rt}$

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So $C(0) = 110$

Husband

Half life of 4 hours. So

$C(4) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{4r}$

$e^{4r} = 0.5$

Applying ln to both sides

$\ln{e^{4r}} = \ln{0.5}$

$4r = \ln{0.5}$

$r = \frac{\ln{0.5}}{4}$

$r = -0.1733$

So for the husband

$C(t) = 110e^{-0.1733t}$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.1733t}$

$C(11) = 110e^{-0.1733*11} = 16.35$

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

$C(10) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{10}$

$e^{10r} = 0.5$

Applying ln to both sides

$\ln{e^{10r}} = \ln{0.5}$

$10r = \ln{0.5}$

$r = \frac{\ln{0.5}}{10}$

$r = -0.0693$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.0693t}$

$C(11) = 110e^{-0.0693*11} = 51.33$

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0

3 years ago