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Simora [160]
3 years ago
12

Solve 3-6 pls and thank you!!

Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

3:810

4:530

I don't know about 5

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How much of a radioactive kind of thorium will be left after 14,680 years if you start with
babymother [125]

Answer:

8978 grams

Step-by-step explanation:

The equation to find the half-life is:

N(t)= N_{0}e^{-kt}

N(t) = amount after the time <em>t</em>

N_{0} = initial amount of substance

t = time

It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.

N(t)= N_{0}e^{-\frac{ln(\frac{1}{2}) }{t_{h} } t} or more simply N(t)= N_{0}(\frac{1}{2})^{\frac{1}{t_{h} } }

t_{h} = time of the half-life

We know that N_{0} = 35,912, t = 14,680, and t_{h}=7,340

Plug these into the equation:

N(t) = 35912(\frac{1}{2})^{\frac{14680}{7340} }

Using a calculator we get:

N(t) = 8978

Therefore, after 14,680 years 8,978 grams of thorium will be left.

Hope this helps!! Ask questions if you need!!

8 0
2 years ago
Picture provided help!
weqwewe [10]

Answer:

A -60i - 14j

Step-by-step explanation:

u = -9i + 8j

v = 7i + 5j

2u = 2(-9i + 8j) = -18i + 16j

6v = 6(7i + 5j) = 42i + 30j

2u - 6v

(-18i + 16j) - (42i + 30j)

(-18i - 42i) + (16j-30j)

-60i - 14j

5 0
3 years ago
Johnny Awesome and his band is raising money to save the world. He is having a concert and charging
Andre45 [30]

Answer:

9375

Step-by-step explanation:

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3 years ago
What is the total area, in square meters, of the shaded sections of the trapezoid below?
max2010maxim [7]

Answer:

84.5

Step-by-step explanation:

(9.9 × 10) ÷ 2= 49.5 (7×10)÷2 = 35

35+49.5 = 84.5

5 0
3 years ago
Beginning with a regular hexagon, a 6-pointed star is formed by constructing equilateral triangles, whose sides are the same len
galina1969 [7]

Answer:

1/12

Step-by-step explanation:

6 0
3 years ago
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