How to find the volume of the object

Mathematics

1 answer:

Serhud [2]3 years ago

6 0

Answer:

measure the length, width and height. Multiply the length times the width and multiply the result by the height. The result is the volume.

Step-by-step explanation:

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How to start and finish

Nitella [24]

$Answer: \\ Since \: it's \: an \: isosceles \: triangle \\ = > x + 15 = 7 \\ < = > x = 7 - 15 = - 8$

8 0

3 years ago

Read 2 more answers

Radioactive Decay

SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$

Percentage of radioactive element is:

Amount after 900 years $=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%$

3 0

3 years ago

You start driving north for 7 miles, turn right, and drive east for another 24 miles. At the end of driving, what is your straig

horsena [70]

Answer:

$AC = 25\ miles$

Step-by-step explanation:

Given:

Distance for north side = 7 miles.

Distance for east side = 24 miles.

We need to find the displacement.

Solution:

Figure shows Point A is starting point and AB = 7 miles is North side distance and BC = 24 miles is east side distance and AC is shown as displacement.

Using Pythagoras theorem to find displacement (AC).

$(AC)^{2}=(AB)^{2}+(BC)^{2}$