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seraphim [82]
3 years ago
5

in a certain pentagon, the interior angles are a,b,c,d,e where a,b,c,d,e are integers strictly less than 180. ("Strictly less th

an 180" means they are "less than and not equal to" 180.) If the median of the interior angles is 61 and there is only one mode, then what are the degree measures of all five angles?
Mathematics
2 answers:
agasfer [191]3 years ago
6 0

Answer:

61,61,61,178,179

Step-by-step explanation:

your welcome. :)

guajiro [1.7K]3 years ago
3 0

Answer:

  least to greatest: {61, 61, 61, 178, 179}

Step-by-step explanation:

If the third-largest angle is 61°, the smallest three angles cannot be larger than 183°. Since the total of all angles must be 540°, and the total of the largest two cannot be greater than 179°×2 = 358°, the sum of the smallest three must be at least 540° -358° = 182°.

So, the possible sets of angles with the smallest 3 totaling 182° or 183° are (in degrees) ...

  {60, 61, 61, 179, 179} . . . . two modes

  (61, 61, 61, 178, 179} . . . . . one mode -- the set you're looking for

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rusak2 [61]

Answer:

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Step-by-step explanation:

This should be your answer if you are doing distributive property.

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Brilliant_brown [7]

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Step-by-step explanation:

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Alex is standing 60 feet from the base of a flagpole. He measures the angle of elevation to the top of the pole as 35º. Vera is
ivanzaharov [21]

Answer:

49.4 degrees

Step-by-step explanation:

In Triangle AXY,

Tan 35^0=\frac{|XY|}{60} \\|XY|=60*Tan 35^0=42.01\:feet\\$Therefore, Height of the pole=42.01 \:feet

We want to determine the angle of elevation from the point Vera is standing to the top of the flagpole, which is the angle at V in the diagram.

In Triangle XVY

|VY|=36 feet

Tan \theta=\frac{|XY|}{|VY|} \\Tan \theta=\frac{42.01}{36}\\ \theta=arctan(\frac{42.01}{36})\\ \theta=49.4^0

Therefore, the angle of elevation from the point Vera is standing to the top of the flagpole is 49.4 degree to the nearest tenth of a degree.

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3 years ago
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Shortly after their arrival, Europeans began introducing pigs to the Americas as a source of food. Some escaped, and others were
natka813 [3]

Answer: About 800 thousand

The more accurate value is 807,528 but this is also an approximation.

=======================================================

Work Shown:

t = \text{number of years since the year 2000}

P_0 = \text{population (in millions) in the year 2000}

P = P_0(1.2)^t\\\\5 = P_0(1.2)^{10}\\\\5 \approx P_0*6.1917364224\\\\P_0 \approx \frac{5}{6.1917364224}\\\\P_0 \approx 0.80752791444922\\\\P_0 \approx 0.807528\\\\

That's the rough population of wild pigs (in millions) for the year 2000.

Multiply by 10^6 to get it in terms of units instead.

0.807528\times10^6 = 807,528

There were roughly 800 thousand wild pigs in the year 2000.

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1 year ago
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