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3 years ago

A CD usually sells for $18.00. If the CD is 30% off, and sales tax is 6%, what is the total price of the CD, including tax?

1 answer:

6 0

Ok first find out what it will cost after the 30% discount

18*.30=
$5.40 discount
$18.00-$5.40=
$12.60
Now lets figure out the sales tax
$12.60*.06=
76 cents
total price including tax is $13.36

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Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

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$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

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$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

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$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

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