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kipiarov [429]
2 years ago
5

The vertices of a quadrilateral drawn in a coordinate plane are (-6,6), (4,6), (2,-5), and (-6,-9). what is the length of the si

de joining the vertex in quadrant i to the vertex in quadrant 11? a) 2 b) 8 c) 10 d) 12​
Mathematics
1 answer:
madreJ [45]2 years ago
8 0

Using the formula for the distance between two points, it is found that the length of the side joining the vertex in quadrant i to the vertex in quadrant ii is given by:

c) 10

<h3>What is the distance between two points?</h3>

Suppose that we have two points, (x_1,y_1) and (x_2,y_2). The distance between them is given by:

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

In quadrant I, both coordinates are positive, hence the vertex is (4,6). In quadrant II, x is negative while y is positive, hence the vertex is (-6,6). Thus, the distance is given by:

D = \sqrt{(-6 - 4)^2+(6 - 6)^2} = 10

Which means that option C is correct.

More can be learned about the distance between two points at brainly.com/question/18345417

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A museum has a wax sculpture of a historical village. The scale is 1.5:8. If the height of a hut in the sculpture is 5 feet, how
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Let us say that x is the real height of the original hut, therefore we can establish the equation:

5 : x = 1.5 : 8

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3 years ago
Which figures demonstrate a translation?
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The two bottom graphs demonstrate translations.

<h3>Which figures demonstrate a translation?</h3>

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If you want to learn more about translations:

brainly.com/question/24850937

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8 0
2 years ago
What's 43080700 in word form
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8 0
3 years ago
What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?
denis23 [38]
The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle  (x,y):
d^2 = (x-x_0)^2 + (y-y_0)^2
Plugging in points A and B into distance formula, then setting them equal to each other gives:
(x-2)^2+(y-6)^2 = (x-2)^2 + y^2
Right away we can cancel out the x terms leaving:
(y-6)^2 = y^2
Expand Left side and Solve for y:
y^2 -12y +36 = y^2
y = 3
Plug in points B and C as before:
(x-2)^2 + y^2 = (x-10)^2 + y^2
Here we can cancel the y-terms.
Expand and solve for x:
x^2 -4x+4 = x^2 -20x+100
16x = 96
x = 6
Therefore the center of the circle is the point (6,3)
6 0
3 years ago
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