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4 years ago

A 3-pack of beaded necklaces costs $4.47. What is the unit price?

Mathematics

2 answers:

inysia [295]4 years ago

7 0

Answer:

1.49 per necklace

Step-by-step explanation:

there are 3 necklaces sold for 4.47. if we divide the three necklaces by the overall price we get the price per necklace. 4.47 divided by 3 = 1.49

stepladder [879]4 years ago

6 0

Answer:

Each necklace cost $1.49

Step-by-step explanation:

Three necklaces cost $4.47. So to see how much one necklace cost you must divide the cost of all three necklaces by 3. Then you get $1.49

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On average, 75% of the days in Henderson county are sunny, with little or no cloud cover. Describe a model that you could use to

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One possible model is:

You could place 3 red marble and 1 blue marble in a bag. The probability of drawing a red marble would be 3/4, which is 75%; this means red marbles would be sunny days and blue marbles would be cloudy days.

Each draw out of the bag would represent one day of the week. Draw a marble 7 times, replacing it after each draw. This would represent the weather for the days of the week.

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3 years ago

Round to the nearest hundred and solve a problem. 372,864+214,209=

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3 years ago

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sdas [7]

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3 years ago

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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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3 years ago

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