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White raven [17]
3 years ago
5

Answer the given question and show workings.

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
8 0

Answer:

  • 1

Step-by-step explanation:

<h3>Given</h3>
  • sin θ = 3/5
  • cos α = 3/5
  • Both angles are acute
<h3>Solution</h3>

Since both sin θ and cos α have same value of 3/5 and both angles are in the I quadrant, the angles θ and α are complementary.

<u>It means: </u>

  • θ + α = 90°
  • sin (θ + α) = sin 90° = 1
Minchanka [31]3 years ago
6 0

Answer:

alpha=cos^-1(0.6)

alpha=53°

sin(37+53)

=1

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12a

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Solve 25n=2×25 for n. You can use a related equation. n = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%2A25%7D%7B25%7D" id="Tex
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A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is us
Soloha48 [4]

Answer:

The Absolute Error is the difference between the actual and measured value.

Absolute \:error = |Actual \:value - Measured \:value|

The Relative Error is the Absolute Error divided by the actual measurement.

Relative \:error = \frac{Absolute \:error}{Actual \:value}

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To find the absolute error and relative error for each measurement made by the glucose monitor you must use the above definitions.

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Absolute \:error = \left|102-104.5\right|\\Absolute \:error =\left|-2.5\right|\\Absolute \:error =2.5

Relative \:error = \frac{2.5}{102.0}=0.0245

b) For a concentration of 96.2 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-96.2\right|\\Absolute \:error =\left|5.8\right|\\Absolute \:error =5.8

Relative \:error = \frac{5.8}{102.0}=0.0569

c) For a concentration of 102.2 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-102.2\right|\\Absolute \:error =\left|-0.2\right|\\Absolute \:error =0.2

Relative \:error = \frac{0.2}{102.0}=0.00196

d) For a concentration of 98.3 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-98.3\right|\\Absolute \:error =\left|3.7\right|\\Absolute \:error =3.7

Relative \:error = \frac{3.7}{102.0}=0.0363

e) For a concentration of 101.8 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-101.8\right|\\Absolute \:error =\left|0.2\right|\\Absolute \:error =0.2

Relative \:error = \frac{0.2}{102.0}=0.00196

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3 years ago
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