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timurjin [86]
1 year ago
13

How tall is the flag pole (please give explanation too)

Mathematics
1 answer:
Tpy6a [65]1 year ago
3 0

Answer:

  22 3/4 ft

Step-by-step explanation:

The diagram seems to show a 6 1/2 ft person casting a 9 ft shadow at the same time a flagpole is casting a 31 1/2 ft shadow. You want the height of the flagpole.

<h3>Proportion</h3>

Shadow lengths are presumed proportional to the height of the object making them. This means ...

  h/(31 1/2) = (6 1/2)/9 . . . . . all lengths in feet

Multiplying by 31 1/2, we get ...

  h = (31 1/2)(6 1/2)/9 = (819/4)/9 = 91/4

  h = 22 3/4 . . . . ft

The flagpole is 22 3/4 feet tall.

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If this was meant to be "divided to" then it would look like this:
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find the parametric equations for the line of intersection of the two planes z = x + y and 5x - y + 2z = 2. Use your equations t
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Answer:

You didn't give the points in which you want the parametric equations be filled, but I have obtained the parametric equations, and they are:

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Step-by-step explanation:

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

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where r_0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

x = ax, y = by, and z = cz

where a, b and c are the coefficients from the vector equation

r = ai + bj + ck

To find the parametric equations for the line of intersection of the planes.

x + y - z = 0

5x - y + 2z = 2

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find v, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are:

For the plane x + y - z = 0, the normal vector is a⟨1, 1, -1⟩

For the plane 5x - y + 2z = 2, the normal vector is b⟨5, -1, 2⟩

The cross product of the normal vectors is

v = a × b =

|i j k|

|1 1 -1|

|5 -1 2|

= i(2 - 1) - j(2 + 5) + k(-1 - 5)

= i - 7j - 6k

v = ⟨1, -7, -6⟩

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set z = 0 in both equations, we get

x + y = 0

5x - y = 2

Adding these equations

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Substituting x = 1/3 back into

x + y = 0

y = -1/3

Putting these values together, the point on the line of intersection is

(1/3, -1/3, 0)

r_0= (1/3) i - (1/3) j + 0 k

r_0​​ = ⟨1/3, -1/3, 0⟩

Now we’ll plug v and r_0​​ into the vector equation.

r = r_0​​ + tv

r = (1/3)i - (1/3)j + 0k + t(i - 7j - 6k)

= (1/3 + t) i - (1/3 + 7t) j - 6t k

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up r = ai + bj + ck with our vector equation,

r = (1/3 + t) i + (-1/3 - 7t) j + (-6t) k

a = (1/3 + t)

b = (-1/3 - 7t)

c = -6t

Therefore, the parametric equations for the line of intersection are

x = (1/3 + t)

y = (-1/3 - 7t)

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