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Gennadij [26K]
3 years ago
9

Solve for Y xy+y+1=y

Mathematics
2 answers:
irakobra [83]3 years ago
8 0
xy+y+1=y\\ \\ xy+1=0\\ \\ xy=-1\\ \\ y=-\frac { 1 }{ x }
Serggg [28]3 years ago
6 0
xy+y+1=y\\
xy+1=0\\
xy=-1\\
y=- \frac{1}{x}

Maja
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Zarrin [17]
The answer to the question

5 0
3 years ago
Find an equation of the circle that has center (-1,1 ) and passes through (1,-6)
Ivahew [28]

Answer:

(x+1)^2+(y-1)^2=53

Step-by-step explanation:

The standard form for the equation of a circle is (x-h)^2+(y-k)^2=r^2, where h is the x coordinate of the center, k is the y coordinate, and r is the radius. To find the radius of this circle, we can simply find the distance from the center to any of the points on its circumference. Using the distance formula, you know that the distance from (-1,1) to (1,-6) is:

\sqrt{(1-(-1))^2+(1-(-6))^2}=\sqrt{2^2+7^2}=\sqrt{4+49}=\sqrt{53}

Therefore, the equation of this circle is:

(x+1)^2+(y-1)^2=53

Hope this helps!

7 0
3 years ago
For what value(s) of k will the relation not be a function?
kolbaska11 [484]

Answer:

  • See below

Step-by-step explanation:

The relation is not a function when x -value is same but y- values are different.

#1

  • k^2 = 4k
  • k = 0, k = 4

#2

  • k^2 - 5k = k + 7
  • k^2 - 6k = 7
  • k^2 - 6k + 9 = 16
  • k - 3 = 4 ⇒ k = 7
  • k - 3 = -4 ⇒ k = -1

#3

  • k^3 - 5k^2 + 3k = -k
  • k^3 - 5k^2 + 4k = 0
  • k(k^2 - 5k + 4) = 0
  • k = 0, k = 1, k = 4

#4

  • |k + 1| + 2 = 8
  • |k + 1| = 6
  • k = 5, k = -7
8 0
3 years ago
5x + 3y=33 when y= 2x + 1
inn [45]
Hope that this helps.

3 0
3 years ago
Pls help me i'll give the brainlist or whatever it is to the first correct answer. pls NO LINKS.
Yakvenalex [24]

Answer:

628 hope it helps

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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