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Deffense [45]
4 years ago
9

Linear functions (with the exception of f(x) = x) can have at most one fixed point. Quadratic functions can have at most two. Fi

nd the fixed points of the function g(x) = x 2 − 12.
Give a quadratic function whose fixed points are x = −2 and x = 3.
Mathematics
1 answer:
lukranit [14]4 years ago
5 0

Answer:

a. x = 3.46 or -3.46. b. x²- x - 6 = 0

Step-by-step explanation:

At the fixed point, g(x) = 0. So, x² - 12 = 0 ⇒ x² = 12 ⇒ x = ±√12

x = 3.46 or -3.46.

The quadratic equation whose fixed points are x = -2 and x = 3. The fixed points are the roots of the quadratic function.

Using x² - (sum of roots)x + product of roots = 0.

x² -(-2 + 3)x + (-2)(3) = 0

x² - (1)x - 6 = 0

x²- x - 6 = 0

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seropon [69]

Answer:

44cm

Step-by-step explanation:

4+4+6+8+8+14 =

 8  +  14 +  22=44

5 0
3 years ago
Please help me on this I don’t know it
allochka39001 [22]

Answer:

180m^2

Step-by-step explanation:

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7 0
3 years ago
What is the equation of the circle with center (1,1) that passes through the point (-3,4)?
MakcuM [25]

Answer:

(x - 1)2 + (y - 1)² = 5²

Step-by-step explanation:

Adapt the standard equation of a circle with center at (h, k) and radius r:

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

Here one point on the circle is (-3, 4), and so the radius is √[(-3)² + 4² ] = 5

Thus, the desired equation is:

(x - 1)^2 + (y - 1)^2 = 5^2      or     (x - 1)2 + (y - 1)² = 5²

3 0
4 years ago
Please help! Offering 15 points with brainiest
Aloiza [94]

Answer:

1/1296

Step-by-step explanation:

P( 5) = number of 5's over total numbers possible

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P( 5,5,5,5) = 1/6 * 1/6 * 1/6* 1/6

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3 0
3 years ago
Read 2 more answers
Find an angle 0 coterminal to 1174°, where 0° Se < 360°.
Diano4ka-milaya [45]

Answer:

94°

Step-by-step explanation:

one way to do the problem is subtract 360 from 1174 and continue subtracting 360 until you come to a number < 360

1174 - 360 = 814

814 - 360  = 454

454 - 360 = 94

8 0
3 years ago
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