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1 year ago

What is the distance between (4, 3) and (7, 3)?

Mathematics

2 answers:

ZanzabumX [31]1 year ago

7 0

There is a difference of 3 on the x axis

iris [78.8K]1 year ago

4 0

For this one 3 is the distance between you’re coordinate points.

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Write an algebraic expression for the perimeter of the swimming pool

Pachacha [2.7K]

Answer: I don't see that you have a diagram of the swimming pool or any given side measures but assuming it is a rectangle, the algebraic expression would be 2(L+W) where L is the length and W is the width.

Step-by-Step Explanation: So when you are finding the perimeter, you are finding the length of the space surrounding something. So you would add up the length of all its sides to find it. In a rectangle, opposite sides are equal, therefore the length will be equal to the side opposite to it and the width will also be equal to the side opposite to it. So if you were to write the expression like that, you would get L+L+W+W, which can be simplified to 2L+2W, or just 2(L+W).

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2 years ago

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2 years ago

Is the ratio of two odd functions odd

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3 years ago

I need help solving this problem

guapka [62]

Answer:

- 5/13

Step-by-step explanation:

tans = 5/12 = height/base

Using Pythagoras, hypotenuse = √12² + 5² = 13

Therefore coss had to be 12/13

But since cos s < 0, cos s = - 12/13

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8 0

2 years ago

Read 2 more answers

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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