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hichkok12 [17]
3 years ago
14

Piece function:

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0

First, make the table so you can sketch the graph:

    x   |    y

    5  |   20        

    4  |    9              x² - 5   ; x > 3

 <u>   3  |    </u><u>4 </u><u>  </u><u><em>(open dot)                      </em></u><em>   ⇒ </em>from the right

  <u>  3  |    </u><u>5</u><u>                  5     ; x = 3       </u>   ⇒ at x = 3

    3  |    2  <em>(open dot)                          ⇒ </em>from the left

    2  |    4              -2x + 8  ; x < 3          

<u>     1   |    6                                           </u>


Next, look at the graph (or table) to find the limits:

lim 3⁺ = 4  <em>as x approaches 3 from the right, y approaches 4</em>

lim 3⁻ = 2  <em>as x approaches 3 from the left, y approaches 2</em>

lim 3 = DNE <em>lim 3⁺ ≠ lim 3⁻ so the limit does not exist</em>

f(3) = 5  <em>when x = 3, y = 5</em>

f(x) is NOT continuous at x = 3 <em>because lim 3 ≠ f(3)</em>

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Answer:

Minor arc BC - 130° (arc at which inscribed angle is subtended)

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Angle BDC is inscribed angle subtended on the arc BC.

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3 years ago
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How do you complete the square of x^2-14x+33=0?
iVinArrow [24]
X² - 14x + 33 = 0 is in the form ax² + bx + c = 0. We will need this for the second step.

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x² - 14x = - 33

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x² - 14x + 49 = - 33 + 49

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Square root both sides
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or
x - 7 = - 4 and x - 7 = 4

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