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agasfer [191]
3 years ago
6

Figure BCDE was transformed using the composition rx-axis ◦ T2,2. If point B on the pre-image lies at (3, 4), what are the coord

inates of B'' on the final image?
Mathematics
2 answers:
serious [3.7K]3 years ago
7 0

Answer:

B (5,-6)

Step-by-step explanation:

____ [38]3 years ago
5 0

Answer:

The answer is "(5, -6)"

Step-by-step explanation:

Given:

The Pre-image line at point B is: (3, 4)

Solution:

\ rx-axis \circ \  T2,2 (3,4) \\\\\Rightarrow  \ rx-axis \ ( T2,2 (3,4) )\\\\ \Rightarrow \ rx-axis ( 5,6  ) \\\\ \Rightarrow  (5, -6)\\

The coordinates of B point is (5,-6).

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Evaluate the expression when b=2 and y=-6.<br><br> -b+47<br><br> Help
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4 0
3 years ago
What is the inverse function of t=5v+20
Alexxx [7]
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6 0
3 years ago
F(x) = x^2 + 4x + 20 find the real roots. take your time if you want :)
s344n2d4d5 [400]

Answer:

No real roots

x = -2 + 4i, x = -2 - 4i

Step-by-step explanation:

Hello!

We can solve the quadratic by using the Quadratic Formula.

Standard form of a Quadratic: ax^2 + bx + c = 0

Quadratic Equation: x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}

Given our Equation: f(x) = x^2 + 4x + 20

  • a = 1
  • b = 4
  • c = 20

Set the equation to 0 and solve using the formula.

<h3>Solve</h3>
  • x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
  • x = \frac{-4\pm\sqrt{4^2 - 4(1)(20)}}{2(1)}
  • x = \frac{-4\pm\sqrt{16 - 80}}{2}
  • x = \frac{-4\pm\sqrt{-64}}{2}
  • x = \frac{-4\pm8i}{2}
  • x = -2 + 4i, x = -2 - 4i

There are no real roots to the quadratic.

8 0
2 years ago
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