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Colt1911 [192]
3 years ago
15

Solve for x: 4 over x equals 5 over 10 8 9 20 40

Mathematics
1 answer:
Sati [7]3 years ago
4 0

\frac{4}{x} =\frac{5}{10}   [4 over x equals 5 over 10]  

Solve for x, so you need to isolate/get x by itself in the equation.

\frac{4}{x} =\frac{5}{10}       Multiply x on both sides

(x)\frac{4}{x}=\frac{5}{10} (x)

4=\frac{5}{10} x       Multiply the inverse of 5/10, which is 10/5, to get rid of the fraction and get x by itself

(\frac{10}{5} )4=(\frac{10}{5} )\frac{5}{10} x

\frac{40}{5} =x

8 = x

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keep in mind that the inverse cosine function has a range of [0, 180°], so any angles it will spit out, will be on either the I quadrant where cosine is positive or the II quadrant, where cosine is negative.

however, 45.6° has a twin, she's at the IV quadrant, where cosine is also positive, and that'd be 360° - 45.6°, or 314.4°.

now, those are the first two, but we have been only working on the [0, 360°] range.... but we can simply go around the circle many times over up to 720° or 72000000000° if we so wish, so let's go just one more time around the circle to find the other fellows.

360° + 45.6° is a full circle and 45.6° more, that will give us the other angle, also in the first quadrant, but after a full cycle, at 405.6°.

then to find her twin on the IV quadrant, we simply keep on going, and that'd be at 360° + 360° - 45.6°, 674.4°.

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6 0
3 years ago
Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows
koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

N(t) = N_{0}e^{-\lambda t}     (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant  

t is the time = 72 d  

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.    

First, we need to find the decay constant:

\lambda = \frac{ln(2)}{t_{1/2}}    (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g    

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.  

I hope it helps you!  

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