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3 years ago

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A Stability cable is attached to the top of an electricity pole and then stretched to the ground.The pole is 4.5m tall while it

he cable has a length of 6m.How far from the base of the pole is the cable attached to the ground

1 answer:

valkas [14]3 years ago

7 0

Answer: 3.96 meters

Step-by-step explanation:

Draw a Right triangle as the one shown in the image attached, where "x" is the distance from the base of the pole to the point in which the cable is attached to the ground.

In this case you need to use the Pythagorean Theorem. According to this theorem:

$a^2=b^2+c^2$

Where "a" is the hypotenuse and "b" and "c" are the legs of the Right triangle.

If you solve for one of the legs, you get:

$a^2-b^2=c^2\\\\c=\sqrt{a^2-b^2}$

In this case, you can identify that:

$a=6\ m\\\\b=4.5\ m\\\\c=x$

Substituting values, you get the following result:

$c=\sqrt{(6\ m)^2-(4.5\ m)^2}\\\\c=3.96\ m$

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A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

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$T_{x}(x,y)=2x-\frac{3}{2}=0$

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The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

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Now, we need to evaluate the boundary condition.

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Now, let's find the critical point again, as we did above.

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Evaluating T(x,y) at this point, we have:

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Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

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