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svet-max [94.6K]

3 years ago

6

A coordinate plane is used to track the movement of a ship. The ship moved from (8, - 4) to (11, 11) on the coordinate plane. Wh

ich of the following translations describes the movement of the ship?

1 answer:

Luba_88 [7]3 years ago

8 0

Answer:

the ship moved 4 coordinates away from the current position

Step-by-step explanation:

because 8-4 is 4

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adoni [48]

Answer:

Area Of a Right Angled∆

=><em><u> </u></em><em><u>1</u></em><em><u>/</u></em><em><u>2</u></em><em><u> </u></em><em><u>x </u></em><em><u>Base </u></em><em><u>x </u></em><em><u>Height</u></em>

Area of a ∆ Using Heron's Formula

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The equation would be

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The worlds two highest dams are both in Tajikistan. The rogue dam is 35 meters taller than the Nurek dam. together they are 635

galina1969 [7]

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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.

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

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

$e^{-6r} = 0.97$

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

$-6r = \ln{0.97}$

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

$r = 0.0051$

So

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

Determine the half-life of the radioactive substance.

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

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

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

$e^{-0.0051t} = 0.5$

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

$-0.0051t = \ln{0.5}$

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

$t = 135.9$

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Dvinal [7]

Answer:

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Step-by-step explanation:

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3 years ago