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Verdich [7]
3 years ago
11

Allegra has a cell-phone plan that charges $65 per month and $0.10 for every minute that she uses the phone beyond what her plan

allows. One month, she was billed $73.40. Define a variable for the unknown. Write an equation to model the problem. Explain your answer. Solve the equation. Show your work. Find the number of minutes that Allegra went over the time that the plan allows. Explain your answer.
Mathematics
2 answers:
Sloan [31]3 years ago
8 0
65+0.1m=bill

bill=73.40

65+0.1m=73.4
minus 65 from both sides
0.1m=8.4
divide both sides by 0.1
m=84
went 84 mins over
Helga [31]3 years ago
3 0
X0.10. xtimesy (y being every minute) = 8.40 (Money that she spent extra) Well 8.40 divided by 0.10 is 84. so she went 84 mins over
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What is negative 30 over twelve
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Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

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B^2 = 100 + 144 - 207.846096907

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Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

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Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

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Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

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Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

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Calculating \theta

Sum of angles in a triangle = 180

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\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

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posledela
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