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zheka24 [161]
1 year ago
8

For the data in the table, write an equation for the direct variation.

Mathematics
1 answer:
melisa1 [442]1 year ago
4 0

The equation of direct variation is,

\begin{gathered} y=kx \\ k=\frac{y}{x} \end{gathered}

Substitute the values of <em>y</em> and <em>x</em>.

\begin{gathered} k=\frac{5.4}{4} \\ k=1.35 \end{gathered}

So, the obtained equation will be <em>y</em>=1.35(x).

Let's check with the values of <em>x</em>.

1) If <em>x</em>=4

\begin{gathered} y=1.35x \\ y=1.35(4) \\ y=5.4 \end{gathered}

2) If <em>x</em>=2

\begin{gathered} y=1.35x \\ y=1.35(2) \\ y=2.7 \end{gathered}

3) If <em>x</em>=-2

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Answer:    \dfrac{1}{2\sqrt x}

<u>Step-by-step explanation:</u>

\lim_{h \to 0} f(x)=\dfrac{f(x+h)-f(x)}{h}

f(x) = \sqrt x\\

f(x+h) = \sqrt{x+h}

\lim_{h \to 0} f(x)=\dfrac{\sqrt{x+h}-\sqrt x}{h}

                   =\dfrac{\sqrt{x+h}-\sqrt x}{h}\bigg(\dfrac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\bigg)

                   =\dfrac{(x + h)-(x)}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{h}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{1}{\sqrt{x+h}+\sqrt x}

                  =\dfrac{1}{\sqrt{x+0}+\sqrt x}

                  =\dfrac{1}{2\sqrt x}

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