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3241004551 [841]
3 years ago
8

1. Refer to the equation 3x − 5y = 15. (a) Create a table of values for at least 4 points. Show your work on how you found the v

alues for each coordinate pair, and validated the points were on the line.
Work for Point 1
Work for Point 2
Work for Point 3
Work for Point 4
Mathematics
2 answers:
MakcuM [25]3 years ago
8 0

Answer: For this problem, you cannot go over 10! so remember guys you can graph it however you want for example if you wanted to you could do -3/5(-2) + 3 for the first point but you would have to solve it. the next one would be maybe -3/5(-1) + 3 if you wanted to go up by plus one REMEMBER this is your OWN graph so do whichever number order you want it doesn't matter you can increase by 1 or 2 or 3 but the final point shouldn't be over ten or you will get the problem wrong. Good Luck I hope I get an A and I hope you guys get A's as well. Do ur absolute best on this

Ann [662]3 years ago
6 0

Answer:

(0, 3 ), (5, 0), (10, -3), (15, -6)

Step-by-step explanation:

3x − 5y = 15

-5y = 3x + 15

y = -\frac{3}{5} x + 3

Point 1:  (0, 3 )

  y = -3/5(0) + 3

  y = 0 + 3

  y = 3

 

  Validate:

   3 = -3/5(0) + 3

   3 = 3

Point 2:  (5, 0)

  y = -3/5(5) + 3

  y = -3 + 3

  y = 0

  Validate:

   0 = -3/5(5) + 3

   0 = 0

Point 3:  (10, -3)

  y = -3/5(10) + 3

  y = -6 + 3

  y = -3

  Validate:

   -3 = -3/5(10) + 3

   -3 = -3

Point 4:  (15, -6)

  y = -3/5(15) + 3

  y = -9 + 3

  y = -6

 Validate:

   -6 = -3/5(15) + 3

   -6 = -6

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As you know that  Centroid is the point of any curve ,Starting from a line segment it's centroid is it's mid point,  want to say that centroid means the point where the curve is balanced such that if at that point if you put anything it will remain balanced in all directions, whether it is a one dimensional figure,two dimensional figure,or three dimensional figure.For a circle it's center is the Centroid.

Now  coming to your question

Centroid of the quarter of the unit circle lying in the third quadrant.

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x^2+y^2=1\\

Applying the rule of integration i.e from (-1,0) to (0,-1). i.e from x_{1}=-1 \text{ to }   x_{2} =0 \text{  and } y_{1}=0\text{ to } y_{2} =-1-----[∵ we have to find centroid of a circle lying in the third quadrant.]

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\bar{y}=\frac{2}{\pi }[x-\frac{x^3}{3}]_{0}^{-1}

\bar{y}=-\frac{4}{3\pi}\\centroid(-\frac{4}{3\pi},-\frac{4}{3\pi})













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