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2 years ago

Help please??????????

Mathematics

1 answer:

marishachu [46]2 years ago

5 0

MachineA: y=120x

MachineA could pack in 12 min
y=120(12)
y=1440

MachineB could pack in 12 min
y=150(12)
y=1800

MachineB pack than MachineA
1800-1440
=360 candies

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You scored 31 out of 40 points on a test. What percent did you get?

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Answer:

Step-by-step explanation:

Divide 31 by 40. Since you get .775, you make it a percent and it's 77.5% correct.

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2 years ago

What is 34 ÷ 14?A) 1B) 3C) 4D) 316

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1 year ago

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What is the measure of the base in the triangle below?

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Step-by-step explanation:

AC+CB+BA=180

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1 year ago

180% is equivalent to what decimal value

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2 years ago

Find all possible values of α+

const2013 [10]

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:

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

recall angle sum indentity of tan:

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

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:

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

likewise by Opposite Angle Identity we obtain:

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

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

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

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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2 years ago

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