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liberstina [14]
3 years ago
9

10(5 - n) - 1 = 29 n = ? Help!

Mathematics
2 answers:
ICE Princess25 [194]3 years ago
6 0

To get rid of the parantheses we must multiply 10 by anything inside the parantheses.

<em>10(5-n)-1=29</em>

<em>50-10n-1=29</em>

Combine like terms

<em>50-10n-1=29</em>

<em>49-10n=29</em>

Subtract 49 from both sides

<em>49-10n=29</em>

<em>-10n=-20</em>

Divide both sides by -10

<em>-10n/-10=-20/-10</em>

<em>n=2</em>

<em>Our final answer is 2</em>

Hope this helps!

nydimaria [60]3 years ago
5 0

n = 2

distribute and simplify left side of equation

50 - 10n - 1 = 29

49 - 10n = 29 ( subtract 49 from both sides )

- 10n = 29 - 49 = - 20

n = \frac{-20}{-10} = 2



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Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

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(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

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r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

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Match the properties with the steps to solving the following equation.

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