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Ad libitum [116K]
3 years ago
10

You are applying fertilizer to a football field. The field is 360 feet long and 160 feet wide. You use 8 pounds of fertilizer pe

r 1,000 square feet. The fertilizer comes in 50-pound bags. How many bags of fertilizer will you need to complete the job?
Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer: 10 bags

Step-by-step explanation:

First we must calculate the field area. Assuming it is a rectangle then its area equals the product of its length times its width

A =360*160\\\\A = 57600 ft^2

You use 8 pounds of fertilizer per 1,000 square feet

So the amount of fertilizer you should use is:

57,600 ft^2 * \frac{8\ pounds}{1,000\ ft^2} = 460.8\ pounds

As the fertilizer comes in bags of 50 pound then you need

460.8\ pounds * \frac{1\ bag}{50\ pounds} = 9.2\ bags

Since you can not buy 9.2 bags then we round up to the next whole number.

Finally you need 10 bags of fertilizer

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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
Kaylis [27]

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:

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

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

r = r_0 + tv

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

5x + x + y - y = 2 + 0

6x = 2

x = 1/3

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)

z = -6t

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Step-by-step explanation:

Not sure about these. You might want to wait for another person to answer

Have a nice day!

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