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lilavasa [31]
2 years ago
15

Geometry question, any help is appreciated

Mathematics
2 answers:
Gekata [30.6K]2 years ago
6 0
Volume ratio      250 : 686 = 125 : 343
Radius ratio       ∛(125) : ∛(343) = 5 : 7  scale factor
Area ratio           (5)² : (7)² = 25 : 49

Scale factor (5:7),  area ratio (25:49), volume ratio (125:343)
kodGreya [7K]2 years ago
4 0

Answer:

Hence, the ratios are given by 5:7, 25:49, 125:343.

Step-by-step explanation:

We are given two spheres such that:

Volume of small sphere V_{s} = 250 yd^{3}

Volume of large sphere V_{l} = 686 yd^{3}.

Then the ratio of volume = 250 : 686 = 125 : 343.

Since, the volume of a sphere = \frac{4 \pi \times r^{3}}{3}, this gives us that the ratio of the radius = ∛V_{s} : ∛V_{l}

i.e. The ratio of the radius = ∛125 : ∛343 = 5 : 7.

Further, as the surface area of a sphere = 4 \pi \times r^{2}, this gives us that the ratioof surface area = r_{s} ^{2} : r_{l} ^{2}

i.e The ratio of the surface area = 5^{2} : 7^{2} = 25 : 49

So, the ratio of the surface areas = 25 : 49.

Hence, the ratios are given by 5:7, 25:49, 125:343.

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If y varies directly with x, and y= 10 when x = 4, find y when x = 10.​
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Step-by-step explanation:

bc

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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li
krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

First, let us calculate the decay constant (k)

75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036

Next, let us calculate the half-life as follows:

\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

Therefore the half-life is between 600 and 700 years

5 0
2 years ago
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
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