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23.06.2022 06:03
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A company purchased a weaving machine for $190,000. The machine has a useful life

A company purchased a weaving machine for $190,000. The machine has a useful life of 8 years and a residual value of $10,000. It is estimated that the machine could produce 75,000 bolts of woven fabric over its useful life. In the first year, 15,000 bolts were produced. In the second year, production increased to 19,000 units. Using the units-of-production method, what is the amount of accumulated depreciation at the end of the second year? A. $48,133.

B. $45,600.

C. $86,133.

D. $23,750.

E. $81,600.
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cobalt3931
cobalt3931
4,6(45 marks)

The greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

Step-by-step explanation:

We can think of each displacement as a vector, with a given magnitude and direction.

A vector can be written using its x and y coordinates like this

\vec{t}=(x, y)

So, for the displacements a and c their vector coordinates would be:

\vec{a}=(2, 0)

\vec{c}=(-1, 0)

As the b displacement has an angle of 30° toward the north from due east, we can find its x and y coordinates using the following formulas:

x=(magnitude)*cos(angle)

y=(magnitude)*sin(angle)

Note: the angle in the formula is the one formed with the east measured counterclockwise.

So, the x and y coordinates for the b displacement will be:

\vec{b}=(2*cos(30), 2*sin(30))

As the vector additon is commutative, the order won't affect the final position. Nevertheless, any change in the direction of any displacement will change the final position. So, in order to find the combination greatest distance we should calculate the following additions and find the one with the greatest magnitude:

\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}

\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}

\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}

\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}

Each resultant vector can be found adding each component. Afterwards, the magnitude can be found using the following formula:

|\vec{R}|=\sqrt[ ]{(R_{x})^2 +{(R_{y})^2}}

Now, let's calculate!

\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}

R_{1_x}} =2+2*cos(30)-1=2.73

R_{1_y}} =0+2*sin(30)+0=1

\vec{R_{1}}=(2.73,1)

|\vec{R_{1}}|=\sqrt[ ]{(2.73)^2 +{(1)^2}}=3.86

\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}

R_{2_x}} =2-2*cos(30)-1=0.73

R_{2_y}} =0-2*sin(30)+0=-1

\vec{R_{2}}=(-0.73,-1)}

|\vec{R_{2}}|=\sqrt[ ]{(-0.73)^2 +{(-1)^2}}=1.03

\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}

R_{3_x}} =2+2*cos(30)+1=4.73

R_{3_y}} =0+2*sin(30)-0=1

\vec{R_{3}}=(4.73,1)

|\vec{R_{3}}|=\sqrt[ ]{(4.73)^2 +{(1)^2}}=6.69

\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}

R_{4_x}} =2-2*cos(30)+1=1.26

R_{4_y}} =0-2*sin(30)-0=1

\vec{R_{4}}=(1.26,-1)

|\vec{R_{4}}|=\sqrt[ ]{(1.26)^2 +{(-1)^2}}=1.79

So, after all the calculation, we can know for sure that the vector \vec{R_{3}} has the biggest magnitude. Then, the greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

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