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3 years ago

What is an equation of the line that passes through the point ( 6 , − 1 ) (6,−1) and is parallel to the line x − 3 y = 3 x−3y=3?

Mathematics

1 answer:

Nataly [62]3 years ago

4 0

Answer:

y = 1

Step-by-step explanation:

Passes through (6,-1)(6,-1)

Find the slope (m)

m=(y2 - y1) / (x2 - x1)

m=(-1 - (-1) ) / (6-6)

m= 0

Parallel to x - 3y = 3

-3y = -x +3

y = 1/3 x -1

Equation of the line equation is y - y1 = m ( x - x1 )

y - ( -1 ) = 0 ( x - 6 )

y + 1 = 0

y = 1

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Verdich [7]

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3 years ago

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The company is considering the purchase of a new machine for $75,660 (based on the available data), and the payback period is <u>24 years</u>.

<h3>What is the payback period?</h3>

The payback period is the time the company requires to recoup its investment for the new machine.

The payback period can be computed by dividing the investment cash outflows by the annual net cash inflows.

<h3>Data and Calculations:</h3>

Initial investment in new machine = $75,660

Annual depreciation expense = $4,600

Investment period = 10 years

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Thus, since the payback period is <u>24 years</u>, while the investment period is 10 years, it sounds unwise for the company to continue the investment.

Learn more about the payback period at brainly.com/question/23149718

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1 year ago

Find the area ratio of a regular octahedron and a tetrahedron regular, knowing that the diagonal of the octahedron is equal to h

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Answer:

$\frac{4}{3}$

Step-by-step explanation:

The area of a regular octahedron is given by:

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area = $2\sqrt{3}\ *a^2$

Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.

a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:

$h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a$

The area of a tetrahedron is given by:

area = $\sqrt{3}\ *edge^2$ = $\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2$

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Ratio = $\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}$

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3 years ago