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lakkis [162]
3 years ago
8

Corinne has a job selling magazines. She earns $7.50 per hour plus 20% of the total amount of her sales. She also gets an allowa

nce of $40 per week for gas. She knows her weekly earnings can be shown using the following expression: 7.50h + 0.20s + 40 Part A: Identify a coefficient, a variable, and a constant in this expression. (3 points) Part B: If Corinne works for 25 hours and sells $300 in magazines, how much does she earn for the week? Show your work to receive full credit. (4 points) Part C: If Corinne gets a raise and begins earning $9 per hour, would the coefficient, variable, or constant in the equation change? Why? (3 points)
Mathematics
1 answer:
Lynna [10]3 years ago
4 0
Part A:  Coefficient:  either 7.50 or 0.20
Part B:

7.50 (25) + 0.20 (300) + 40
187.50 + 60 + 40 = $287.50

Part C: After her raise, the first term will change from 7.50h to 9h
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Please show the working and answer. you can take a picture for the working.
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Answer:

(a) The area of the triangle is approximately 39.0223 cm²

(b) ∠SQR is approximately 55.582°

Step-by-step explanation:

(a) By sin rule, we have;

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5.4/(sin(52°)) = 6.8/(sin(∠PSQ))

∴ (sin(∠PSQ)) = (6.8/5.4) × (sin(52°)) ≈ 0.9923

∠PSQ = sin⁻¹(0.9923) ≈ 82.88976°

Similarly, we have;

5.4/(sin(52°)) = SP/(sin(180 - 52 - 82.88976))

Where, 180 - 52 - 82.88976 = ∠PQS = 45.11024

SP = 5.4/(sin(52°))×(sin(180 - 52 - 82.88976)) ≈ 4.8549

Given that RS : SP = 2 : 1, we have;

RS = 2 × SP = 2 × 4.8549 ≈ 9.7098

We have by cosine rule, \overline {RQ}² =  \overline {SQ}² +  \overline {SR}² - 2 × \overline {SQ} × \overline {SR} × cos(∠QSR)

∠QSR and ∠PSQ are supplementary angles, therefore;

∠QSR = 180° - ∠PSQ = 180° - 82.88976° = 97.11024°

∠QSR = 97.11024°

Therefore;

\overline {RQ}² =  5.4² +  9.7098² - 2 ×  5.4×9.7098× cos(97.11024)

\overline {RQ}² ≈ 136.42

\overline {RQ} = √(136.42) ≈ 11.6799

The area of the triangle = 1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

By substituting the values, we have;

1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

1/2 × 6.8 × (4.8549 + 9.7098) × sin(52°) ≈ 39.0223 cm²

The area of the triangle ≈ 39.0223 cm²

(b) By sin rule, we have;

\overline {RS}/(sin(∠SQR)) = \overline {RQ}/(sin(∠QSR))

By substituting, we have;

9.7098/(sin(∠SQR)) = 11.6799/(sin(97.11024))

sin(∠SQR) = 9.7098/(11.6799/(sin(97.11024))) ≈ 0.82493

∠SQR = sin⁻¹(0.82493) ≈ 55.582°.

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