All categories

3 years ago

Roger is paid $50 to sell videos of a play to an audience after the play is preformed.

Mathematics

1 answer:

konstantin123 [22]3 years ago

3 0

Answer:

Roger can earn $510 at most.

Step-by-step explanation:

We are given the function

Which gives the earnings of Roger when he sells v videos. Since the play’s audience consists of 230 people and each one buys no more than one video, v can take values from 0 to 230, i.e.

That is the practical domain of E(v)

If Roger is in bad luck and nobody is willing to purchase a video, v=0

If Roger is in a perfectly lucky night and every person from the audience wants to purchase a video, then v=230. It's the practical upper limit since each person can only purchase 1 video

In the above-mentioned case, where v=230, then

Roger can earn $510 at most.

You might be interested in

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago

4 1/2 + 1 + 2 + 1 1/2 + 4 + 1/2

mafiozo [28]

Your answer would be 17 1/2

7 0

3 years ago

Can someone help me with 5-6

olchik [2.2K]

Answer:

B, C

Step-by-step explanation:

I love slope

4 0

3 years ago

G C D of 130 and 150

wlad13 [49]

Answer:

GCF = 10

Step-by-step explanation:

8 0

2 years ago

This math I just can't figure out

Elena-2011 [213]

Step-by-step explanation:

Here's the first question and how to solve the second.

So the first business start with 250, that's your b in slope.

y = mx + b

y = mx + 250

Now, they charge 14 an hour so that's your m.

y = 14x + 250

x is the number of hours it takes.

The second business is straight up 22 per hour, so it's just y = mx, where b is 0.

y = 22x VS y = 14x + 250

Try plugging in different numbers and seeing how much it take to level out!

4 0

3 years ago