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kotykmax [81]
3 years ago
12

a line parallel to a triangle’s side splits line AB into lengths of x-6 and x. the other side, line AC, is split into lengths of

x+6 and x+20. what is the length of AC ?
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
6 0

Answer:

56

Step-by-step explanation:

Let's say that D is the point where AB is split, so AD = x-6 and DB = x.

And let's say that E is the point where AC is split, so AE = x+6 and EC = x+20.

Triangle ADE is similar to triangle ABC.  Therefore:

(x-6) / (x-6 + x) = (x+6) / (x+6 + x+20)

Solving:

(x-6) / (2x-6) = (x+6) / (2x+26)

(x-6) (2x+26) = (x+6) (2x-6)

2x² + 26x - 12x - 156 = 2x² - 6x + 12x - 36

2x² + 14x - 156 = 2x² + 6x - 36

14x - 156 = 6x - 36

8x = 120

x = 15

So the length of AC is:

AC = x+6 + x+20

AC = 2x + 26

AC = 2(15) + 26

AC = 56

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Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

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The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

\Delta V = 4\pi \cdot r^{2}\cdot \Delta r

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

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Answer:

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Step-by-step explanation:

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