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Alex Ar [27]
3 years ago
6

Jamie needs to build a fence around his garden, as illustrated by polygon ABCDEF on the coordinate grid below. If each unit repr

esents one yard, what is the total length of Jamie's fence in yards?
(72 POINTS)
Please don't just answer for points if you know the answer you know the answer

(Homework assignment worth a test grade so need someone who knows their stuff)

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer:

C. 26 yards

Step-by-step explanation:

Jamie's fence total length = perimeter of the polygon

Perimeter of the polygon = AB + BC + CD + DE + EF + FA

AB, FA and DE can be worked accordingly as shown below:

AB = |-5 - 0| = 5 units

FA = |5 - 2| = 3 units

DE = |1 -(-2)| = 3 units

BC, CD, and EF can be calculated using the formula d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between B(0, 5) and C(4, 2):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(0, 5) = (x_1, y_1)

C(4, 2) = (x_2, y_2)

BC = \sqrt{(4 - 0)^2 + (2 - 5)^2}

BC = \sqrt{(4)^2 + (-3)^2}

BC = \sqrt{16 + 9} = \sqrt{25}

BC = 5 units

Distance between C(4, 2) and D(1, -2)

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(4, 2) = (x_1, y_1)

D(1, -2) = (x_2, y_2)

CD = \sqrt{(1 - 4)^2 + (-2 - 2)^2}

CD = \sqrt{(-3)^2 + (-4)^2}

CD = \sqrt{9 + 16} = \sqrt{25}

CD = 5 units

Distance between E(-2, -2) and F(-5, 2):

EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

E(-2, -2) = (x_1, y_1)

F(-5, 2) = (x_2, y_2)

EF = \sqrt{(-5 -(-2))^2 + (2 -(-2))^2}

EF = \sqrt{(-3)^2 + (4)^2}

EF = \sqrt{9 + 16} = \sqrt{25}

EF = 5 units

Total length of the wall in yards = 5 + 5 + 5 + 3 + 5 + 3 = 26 yards

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

8 0
3 years ago
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