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Artist 52 [7]
3 years ago
6

In David’s class, the ratio of right-handed students to left-handed students is 24 to 6. How many right-handed students are ther

e for each left-handed?
Mathematics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

There are 4 right handed students for each left hand because, you just divide 24 by 6 and you get 4 for each of the 6 left handed students.

Step-by-step explanation:

Or just group, by doing 6 circles and just put one in each of them so then you can see when all of the circles are taken up .

Thanks! Please Mark Brainiest! <3

~   BiologyisAmazing

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The diagram is below

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  • B = (2,1)
  • C = (0,3)
  • D = (-2,1)

Let's find the distance from A to B using the distance formula.

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-2)^2 + (-1-1)^2}\\\\d = \sqrt{(-2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\d \approx 2.828\\\\

The distance from A to B is exactly \sqrt{8} units or approximately 2.828 units. This is the length of side AB.

Through very similar steps, you should find that the other side lengths of BC, CD and AD are all the same.

Since AB = BC = CD = AD, this means we have a rhombus. Recall a rhombus is any quadrilateral with all four sides the same length.

-----------------------------

Let's find the slope of line AB

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{1 - (-1)}{2 - 0}\\\\m = \frac{1 + 1}{2 - 0}\\\\m = \frac{2}{2}\\\\m = 1\\\\

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Rule: Whenever two slopes multiply to -1, then their corresponding lines are perpendicular.

Based on that rule, lines AB and BC are perpendicular. This means they meet up at a right angle (aka 90 degree angle).

You should find that line CD has a slope of 1, so BC and CD are perpendicular. Lastly, line AD has a slope of -1. So all the slopes mentioned are either +1 or -1. The slopes alternate as you move around the figure. This all shows that any two adjacent sides meet at a 90 degree angle.

Therefore, this figure is a rectangle. Any rectangle has all four angles of 90 degrees.

-----------------------------

The first section showed that we have a rhombus. The second section afterward showed that we also have a rectangle. The combination of a rhombus and a rectangle at the same time leads to a square.

Therefore, quadrilateral ABCD is a square.

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