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Leokris [45]
3 years ago
6

The centroid of a triangle is found by constructing the _____.

Mathematics
2 answers:
madreJ [45]3 years ago
8 0
<span>The centroid of a triangle is found by constructing the medians </span>
kkurt [141]3 years ago
6 0

Answer:

<u><em>Medians</em></u>

Step-by-step explanation:

The point where the intersection of the medians that are part of a triangle occurs is called the centroid or barycenter. A median is a line that is drawn from one vertex of the figure to the point that is in the middle of the opposite side. It should be note that the distance between the centroid and its corresponding vertex is twice the distance between the centroid and the opposite side. That is, the distance from the centroid to each vertex is 2/3 the length of each median.

I leave you an image where you can see what is the centroid of a triangle.

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Brainliest and Five Stars for correct answers
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3 years ago
A person in a casino decides to play 3 games of blackjack. let w denote a win and l denote a loss. define the event a as "the pe
Ann [662]

Answer: The answer is {(w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l), (l, w, w)}.



Step-by-step explanation: Given that a person in the casino is going to play 3 games of blackjack. Here, 'w' denotes a win and 'l' denotes a loss.  

Also, let 'A' be the event that the person wins at least one game of blackjack. Then, the outcomes of Event 'A' are

(i) win in 1st game     loss in 2nd game       loss in 3rd game

(ii) win in 1st game     win in 2nd game        loss in 3rd game

(iii) win in 1st game     loss in 2nd game       win in 3rd game

(iv) win in 1st game     win in 2nd game        win in 3rd game

(v) loss in 1st game     loss in 2nd game       win in 3rd game

(vi) loss in 1st game     win in 2nd game        loss in 3rd game

(vii) loss in 1st game     win in 2nd game       win in 3rd game

Thus, the outcomes of the event A are (w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l) and (l, w, w).

Thus, A = {(w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l), (l, w, w)}.


5 0
3 years ago
Maurice and Johanna have appreciated the help you have provided them and their company Pythgo-grass. They have decided to let yo
jasenka [17]
<span><span>1.A triangular section of a lawn will be converted to river rock instead of grass. Maurice insists that the only way to find a missing side length is to use the Law of Cosines. Johanna exclaims that only the Law of Sines will be useful. Describe a scenario where Maurice is correct, a scenario where Johanna is correct, and a scenario where both laws are able to be used. Use complete sentences and example measurements when necessary.
</span>
The Law of Cosines is always preferable when there's a choice.  There will be two triangle angles (between 0 and 180 degrees) that share the same sine (supplementary angles) but the value of the cosine uniquely determines a triangle angle.

To find a missing side, we use the Law of Cosines when we know two sides and their included angle.   We use the Law of Sines when we know another side and all the triangle angles.  (We only need to know two of three to know all three, because they add to 180.  There are only two degrees of freedom, to answer a different question I just did.

<span>2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.
</span>
We'll use the standard parabola, oriented in the usual way.  In that case the directrix is a line y=k and the focus is a point (p,q).

The points (x,y) on the parabola are equidistant from the line to the point.  Since the distances are equal so are the squared distances.

The squared distance from (x,y) to the line y=k is </span>(y-k)^2
<span>
The squared distance from (x,y) to (p,q) is </span>(x-p)^2+(y-q)^2.<span>
These are equal in a parabola:

</span>
(y-k)^2 =(x-p)^2+(y-q)^2<span>

</span>y^2-2ky + k^2 =(x-p)^2+y^2-2qy + q^2

y^2-2ky + k^2 =(x-p)^2 + y^2 - 2qy+ q^2

2(q-k)y =(x-p)^2+ q^2-k^2

y = \dfrac{1}{2(q-k)} ( (x-p)^2+ q^2-k^2)

Gotta go; more later if I can.

<span>3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.

5.Two pillars have been delivered for the support of a shade structure in the backyard. They are both ten feet tall and the cross-sections​ of each pillar have the same area. Explain how you know these pillars have the same volume without knowing whether the pillars are the same shape.</span>
3 0
2 years ago
Sheryl's mean average on eight exams is 92.000. Find the sum of her scores.
storchak [24]
D) 736

8 exams = 92.000
(8)x/8 = 92.000(8)
x = 736

736
7 0
3 years ago
6/16 plus 6/40 plus 6/15
kirza4 [7]
The answer for this question is 37/40
4 0
3 years ago
Read 2 more answers
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