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myrzilka [38]
3 years ago
15

What the image of (- 6, 5) after a reflection over the line y = x ?

Mathematics
1 answer:
zubka84 [21]3 years ago
7 0

Answer:

25

Step-by-step explanation:

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The equation (x - h)^2 + (y - k)^2 = r^2? defines a circle with radius r and center (h, k). Find the center and radius AND graph
bazaltina [42]

9514 1404 393

Answer:

  • center: (1, -3)
  • radius: 5

Step-by-step explanation:

Comparing the two given equations, we see ...

  h = 1

  k = -3

  r = √25 = 5

The center is (1, -3), and the radius is 5.

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3 years ago
4) Determine if the following situation is arithmetic or geometric. Then answer the question using the formula.Eric calls 4 peop
natima [27]

it is a geometric situation and the formula is the following

P(t)=4^t

so after 6 hours we get

P(6)=4^6=4096

8 0
1 year ago
Dante it by 15 pounds of metal for $36 which purchase would have the same cost per pound is Dante purchase
Lostsunrise [7]
36 dollars worth of metal
6 0
2 years ago
17 - 2n = 9 what’s the answer ?
BlackZzzverrR [31]

Step 1: Simplify both sides of the equation.

17−2n=9

17+−2n=9

−2n+17=9

Step 2: Subtract 17 from both sides.

−2n+17−17=9−17

−2n=−8

Step 3: Divide both sides by -2.

−2n−2=−8−2

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8 0
3 years ago
X-y=5 and x^2y=5x+6​
sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5      (1)

x² · y = 5 · x + 6       (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

To learn more on nonlinear equations: brainly.com/question/20242917

#SPJ1

8 0
1 year ago
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