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vitfil [10]
2 years ago
13

Maia is comparing the costs for one month's usage of two cell phone companies. Company A charges $\$40$ per month plus $\$8.50$

per gigabyte of data used. Company B charges $\$30$ per month plus $\$10.50$ per gigabyte of data used.
Mathematics
1 answer:
lesya692 [45]2 years ago
6 0

Answer:

5 is the answer

Step-by-step explanation:

This is not my solution but thought you could use it!

Answer:

company A:  y= 40+8.5*(x)

company B: y=30+10.5*(x)

5 gigabytes of data would have to be used for company A and B to cost the same amount.

Step-by-step explanation:

let the cost be represented on y axis. so, the equation for company A is:

(for the first month) : y= 40+8.5*(x)

[common monthly charges are $40 and let x represent the number of gigabytes of data used. so, 8.5*x gives cost due to to total gigabytes. hence, i have added 40 and  8.5*x to represent the total cost for first month]

company B:

y=30+10.5*(x)

[obtain this also in the above mentioned manner]

the graph is uploaded in the attachment.

to know about number of gigabytes required to make the costs of both the company equal, we can directly observe in the graph where both the company's lines intersect and the get the corresponding x coordinate. this x coordinate represents the required number of gigabytes. or we can directly equate the above equations.[as shown below]

40+8.5*(x)=30+10.5*(x)

subtracting 30 from both sides,

10+8.5x=10.5x

subtracting 8.5x from both sides,

10=2x

dividing both the sides by 2.5,

5=x [since, 10/2 = 5)

therefore, 5 gigabytes of data would have to be used for company A and B to cost the same amount.

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Novay_Z [31]

Answer:

$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $

$ \frac{-(\sqrt{3} + 1)}{2\sqrt{2}} $

$ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $

Step-by-step explanation:

Given $ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $

(A) $ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4}  + \frac{\pi}{6}) $

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $

$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

(B) Cos(A + B) = CosAcosB - SinASinB

$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $

$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $

$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $

$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}

(C) Tan(A + B) = $ \frac{Sin(A +B)}{Cos(A + B)} $

From the above obtained values this can be calculated and the value is $ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $.

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