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3 years ago

What is the distance between (-6, 2) and (8, 10) on a coordinate grid?

Mathematics

1 answer:

lawyer [7]3 years ago

5 0

Answer:

The answer is 2(the square root of 65 )

Step-by-step explanation:

Plug ( -6, 2) and (8, 10) into the distance formula and solve.

Distance = the square root of ((x2 - x1)^2 + (y2 - y1)^2))

Distance = the square root of ((8 - -6)^2 + (10 - 2)^2))

Distance = the square root of (14^2 + 8^2)

Distance = the square root of (196 + 64)

Distance = the square root of 260 which can be simplified to 2(the square root of 64)

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maxonik [38]

Answer:

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Step-by-step explanation:

This is very simple, but I can help more if you're stuck.

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2 years ago

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7nadin3 [17]

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Answer:

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Step-by-step explanation:

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2 years ago

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lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

The Derivative of a Function

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

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Where u is a function of x as provided:

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If we set

$u=(x+\sqrt{1+x^2})$

Then

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$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

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Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

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Lina20 [59]

Answer:

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Step-by-step explanation:

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