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In-s [12.5K]
3 years ago
8

Can someone help me please because I have a lot of trouble doing these problems. Please show work so I can see how you did it.

Mathematics
2 answers:
m_a_m_a [10]3 years ago
4 0
Your answer would be 14

14 + (-9) - (-9)
so (-9) - (-9) would be zero since we are subtracting a negative
14+0 = 14

I hope this helps :)
Juliette [100K]3 years ago
3 0
14+(-9)-(-9)=14-9+9=14
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Any graph will represent a function if it passes the vertical line test. A graph is a function if for any vertical line drawn it touches the graph at only one point.In other  words for every x there is only one y value.

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Element X decays radioactively with a half life of 12 minutes. If there are 160
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It would take 75.8 minutes for the element to decay to 2 grams.

Step-by-step explanation:

The number of grams of element x, after t minutes, is given by the following equation:

X(t) = X(0)(1-r)^{t}

In which X(0) is the initial amount and r is the decay rate.

There are 160 grams of Element X

This means that X(0) = 160.

So

X(t) = X(0)(1-r)^{t}

X(t) = 160(1-r)^{t}

Half life of 12 minutes.

This means that X(12) = 0.5*X(0) = 0.5*160 = 80. So

X(t) = 160(1-r)^{t}

80 = 160(1-r)^{12}

(1 - r)^{12} = 0.5

\sqrt[12]{(1 - r)^{12}} = \sqrt[12]{0.5}

1 - r = 0.9438

So

X(t) = 160(1-r)^{t}

X(t) = 160(0.9438)^{t}

How long would it take for the element to decay to 2 grams?

This is t for which X(t) = 2. So

X(t) = 160(0.9438)^{t}

2 = 160(0.9438)^{t}

(0.9438)^{t} = \frac{2}{160}

\log{(0.9438)^{t}} = \log{\frac{2}{160}}

t\log{0.9438} = \log{\frac{2}{160}}

t = \frac{\log{\frac{2}{160}}}{\log{0.9438}}

t = 75.8

It would take 75.8 minutes for the element to decay to 2 grams.

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3 years ago
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