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Mathematics

1 answer:

Lilit [14]1 year ago

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The C is perpendicular to AB

When using paper, the paper perpendicular te construct to is folded so that AB is divided equally and the two sides of. line AB will lie exactly on top of each other.

When using a reflective device to construct a live perpendicular to line AB, set One end of the device to coincide with point C, and then swing the other end of the device so that the reflection of line AB coincides with the midpoint of AB

The question is incomplete hence the complete question is shown here

Select the correct answer from the drop-down menu. Compare the steps for constructing a perpendicular line to line through point using a reflective device and using paper. , When using paper to construct the perpendicular line, the paper is folded so that and the two sides of line will lie exactly on top of each other. When using a reflective dewice to construct a line perpendicular to line , set one end of the device to coincide with , and then swing the other end of the device so the reflection of line coincides with.

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How do I find dy/dx of the following?

Pachacha [2.7K]

Answer:

$\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$

Step-by-step explanation:

$y \ = \ x^{3} \ - \ \displaystyle\frac{1}{\sqrt{x}} \ + \ \displaystyle\frac{3}{x^{4}} \\ \\ y \ = \ x^{3} \ - \ x^{-\frac{1}{2}} \ + \ 3x^{-4} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{3 \ - \ 1} \ - \ \left(-\displaystyle\frac{1}{2}\right)x^{-\frac{1}{2} \ - \ 1} \ + \ \left(-4 \ \times \ 3\right)x^{-4-1}$

$\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2}x^{-\frac{3}{2}} \ - \ 12x^{-5} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$