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3 years ago

Find the inverse of the equation y=2/3x-6

Mathematics

1 answer:

Lisa [10]3 years ago

4 0

Answer:

2/3x−6

Step-by-step explanation:

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3 years ago

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3 years ago

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AlladinOne [14]

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3 years ago

A certain triangle has two 45° angles. What type of triangle is it?

Lady bird [3.3K]

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3 years ago

Read 2 more answers

A rectangle has its vertices at (-4, -3), (-4,7), (1,7), (1, -3). What part, in percent, of the rectangle is located in Quadrant

Rasek [7]

Consider the attached figure. The whole rectangle is ABCD, while AEGF is the part located in the third quadrant. In fact, this quadrant is composed by all the points with both coordinates negative.

To answer the question, let's compute the area of the two rectangles and see what part of ABCD is AEGF.

A and B have the same x coordinate, so the length of AB is given by the absolute difference of their y coordinates:

$\overline{AB} = |A_y-B_y| = |-3-7| = |-10| = 10$

Similarly, but exchanging the role of x and y, we compute the length of BC:

$\overline{BC} = |B_x-C_x| = |-4-1| = |-5| = 5$

So, the area of the rectangle is $\overline{AB} \cdot \overline{BC} = 10\cdot 5 = 50$

The same procedure allows us to compute width and height of the sub-rectangle in the third quadrant:

$\overline{AE} = |A_y-E_y| = |-3-0| = |-3| = 3$

$\overline{EG} = |E_x-G_x| = |-4-0| = |-4| = 4$

So, the area of the portion located in the third quadrant is $\overline{AE} \cdot \overline{EG} = 3\cdot 4 = 12$

This means that the ratio between the two area is

$\cfrac{\text{area }AEGF}{\text{area }ABCD} = \cfrac{12}{50}$

If we want this ratio to be a percentage, just make sure that the denominator is 100:

$\cfrac{12}{50} = \cfrac{12}{50}\cdot \cfrac{2}{2} = \cfrac{24}{100} = 24\%$

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3 years ago