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mr Goodwill [35]
2 years ago
10

danny is installing a fence around his rectangular yard. His yard is 20feet by 45feet. If the fence cost 25 per foot based on th

e area what is the cost?
Mathematics
2 answers:
3241004551 [841]2 years ago
6 0

Answer:

$225

Step-by-step explanation:

The area to be fenced in is 20 ft by 45 ft, or 900 ft^2.  But "25 per foot based on the area" is illogical.  Please ensure that you have copied this problem down accurately, and be sure to provide units of measurement:  did you mean $25 per square foot, or per foot, or what?

The fencing cost is usually dependent on the unit cost and the length of the yard perimeter.  The perimeter is P = 2L + 2W, which here is P = 2(45 ft) + 2(20 ft), or 130 ft.  If the fencing is 25 cents per foot, $130 ft of fencing would come to $32.50.

               

kolezko [41]2 years ago
6 0

Answer: Find the perimeter of the yard ( the distance around the yard).

A rectangle has two sides of each dimension.

The perimeter is 20 + 20 + 45 + 45 = 130 feet.

Now multiply the total feet by the cost per foot:

130 feet x 25 per foot = $3,250 total.

Step-by-step explanation:

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Answer:

The distance between (-1,5) and (2,5) is √9 or 3.

Step-by-step explanation:

Okay first, just for explaining purposes, let's call the ordered pair (-1,5) letter A and ordered pair (2,5) letter B.

Now to the answer. To start you use the distance formula. which is:

d=√( (x2-x1)²+(y2-y1)²)

Then you solve by plugging the ordered pairs into the formula where -1 from point A is X1 and 5 from point A is Y1. 2 from point B is X2 and 5 from point B is Y2.

When you solve it should look like this:

1. d=√( (x2-x1)²+(y2-y1)²)

2. d=√( (2- -1)²+(5-5)²) *2 minus a negative one means plus so we get 3*

3.  d=√( (3)²+(0)²)

4. d=√( 9+0)

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Evaluate \vec F at \vec r :

\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k

Compute the line element d\vec r :

d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt

Simplifying the integrand, we have

\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt

Then the line integral evaluates to

\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}

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