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2 years ago

Write the equation of the given line in standard form.

Mathematics

1 answer:

ollegr [7]2 years ago

7 0

let's notice that the x-coordinate is the same for both points.

Check the picture below.

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

2 years ago

A stand was selling 8 mangos for 10 dollars at a farmer's market.

MaRussiya [10]

(Ms)Kisha is correct.

How?

Well, I figured this out by using common sense. 8 mangoes are 10 dollars. Since the price is higher than the amount, so it has to be at least more than one.

We could write more ratios. You can always write more ratios. You just multiply/divide both sides. Examples below:
2:5
16:20

5 0

2 years ago

Solve. T + 23 > 1 t > 24 t > –22 t < 24 t < –22

creativ13 [48]

t + 23 > 1 subtract 23 from both sides

t + 23 - 23 > 1 - 23

t > -22