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DaniilM [7]
3 years ago
13

a line is parallel to y=3x-12 and intersects the point (1,-2) what is the equation of this parallel line

Mathematics
1 answer:
nekit [7.7K]3 years ago
7 0
1=3 x -2 -12 = -18 i believe:)
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Franco begins to save for a new Lego set that costs $65. His parents gave him $35 towards the cost
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QUOTIENT OF 84 AND 6
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vodka [1.7K]
You could apply a clever Algebra trick to avoid using the quotient rule,

\rm y=\dfrac{log(x)-1+1}{log(x)-1}=\dfrac{log(x)-1}{log(x)-1}+\dfrac{1}{log(x)-1}

\rm y=1+\dfrac{1}{log(x)-1}

and apply power rule into chain rule from that point.
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Let's start by "setting up" our quotient rule:

\rm y'=\dfrac{(log x)'(log x-1)-log x(log x-1)'}{(log x-1)^2}

If this log notation is not intended to be natural log, then we'll have a little bit of extra work. Our change of base identity allows us to rewrite log base 10 in terms of the natural log,

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so let's apply this to our problem,

\rm y'=\dfrac{\left(\frac{ln x}{ln 10}\right)'(log x-1)-log x\left(\dfrac{ln x}{ln 10}-1\right)'}{(log x-1)^2}

Derivative of ln(x) gives us 1/x in each case,

\rm y'=\dfrac{\left(\frac{1}{x ln 10}\right)(log x-1)-log x\left(\dfrac{1}{x ln 10}\right)}{(log x-1)^2}

Factor the 1/(x ln10) out of each term in the numerator,

\rm y'=\left(\dfrac{1}{x ln 10}\right)\dfrac{log x-1-log x}{(log x-1)^2}

and combine like-terms,

\rm y'=\dfrac{-1}{x(log x-1)^2ln 10}

Lemme know if you're confused with any of the steps.
6 0
4 years ago
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