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fenix001 [56]
3 years ago
6

PLEASE HELP! Last question! Which graphs show the solution to the given inequalities?

Mathematics
1 answer:
charle [14.2K]3 years ago
7 0
1. A, It goes past the 4, but the circle is not shaded because the answer cannot be 4
2. C. It goes below the 11 and is shaded because it can be equal to 11
3. D. It is less than 13 and cannot equal thirteen, hence the non shaded circle
4. B. I goes past the -2 and has the circle shaded because it can be equal to the -2
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\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}

General Formulas and Concepts:  <u> </u>

<u>Algebra I</u>  

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Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle y = \sqrt{x}(8x - 5)

<u>Step 2: Differentiate</u>

\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)

  1. Product Rule:                                                                                                  \displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]
  2. Rewrite:                                                                                                           \displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]
  3. Basic Power Rule:                                                                                          \displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}
  4. Simplify:                                                                                                          \displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}
  5. Rewrite:                                                                                                           \displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8
  6. Multiply:                                                                                                           \displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}
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