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2 years ago

40 POINTS !!! What is the extraneous solution (undefined) to the equation

Mathematics

2 answers:

aleksklad [387]2 years ago

5 0

Answer:

x = 4

x= -6

(I apologize if it is wrong.)

NikAS [45]2 years ago

4 0

X should equal -6 and 4 but 4 doesn’t work.

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goblinko [34]

Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

What is the slope of the line that passes through the points (1, -6)(1,−6) and (-8, 9) ?(−8,9)?

Monica [59]

Answer:

The answer to your question is m = -5/3

Step-by-step explanation:

Data

Point 1 = (1, -6)

Point 2 = (-8, 9)

Process

To find the slope just substitute the values in the formula and simplify.

Formula

slope = m = (y2 - y1) / (x2 - x1)

-Identify x1, y1, x2 and y2

x1 = 1 y1 = -6 x2 = -8 y2 = 9

-Substitution

m = (9 + 6) / (-8 - 1)

-Simplification

m = 15/-9

-Result

m = -5/3

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3 years ago

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3 years ago

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Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3

Whitepunk [10]

If $f(x)=4x-3$ :

$\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4$

If $f(x)=4x^{-3}$ :

$\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}$

$\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}$

$\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}$

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4 years ago