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zmey [24]
3 years ago
15

How to find length of a triangle given one side and angle equladera; triangle?

Mathematics
1 answer:
serious [3.7K]3 years ago
4 0
An equilateral triangle is a triangle with all equal sides, meaning, if the problem gives you one side, then it has given the length of all the sides because all sides lengths are equal. Given that you have one side and one angle, all sides will be the same length and no matter what all angles are the same and if correct the angles will be equal to 60 each. 
I hope this helps some.
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Semenov [28]

Answer:

  $1769.06

Step-by-step explanation:

The sum of Mia's three highest salaries is ...

  $92,000 +94,800 +96,250 = $283,050

so the average is ...

  $283,050/3 = $94,350

Her pension is then ...

  0.01875 × $94,350 = $1769.06

_____

<em>Additional comment</em>

This is equivalent to the total of those salaries, divided by 160.

  T/3 × 0.01875 = T × (0.01875/3) = T × 0.00625 = T/160

Sometimes, there are simpler ways to calculate things like this.

8 0
3 years ago
Rewrite in simplest terms. 4 (-6g - 10h) + 3h - 4 (- 3h - 10g)
kondor19780726 [428]

Answer: 16g - 25h

4 (-6g - 10h) + 3h - 4 (- 3h - 10g)

-24g - 40h +3h +12h +40g

16g - 25h

Step-by-step explanation:

if this is wrong im sorry

8 0
3 years ago
Natalie drives from her home to the art museum. She knows that when she passes the fire station,
scoundrel [369]

Answer:

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Step-by-step explanation:

7 0
2 years ago
Seven-eighths equals 0.875 and is a(n) ___.
kirza4 [7]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No
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If A and B are independent, then P(A\cap B)=P(A)P(B).

a.

P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)

P(A\cup B)=0.5+0.2-0.5\cdot0.2

\boxed{P(A\cup B)=0.6}

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)

P(A^c\cap B^c)=1-0.6

\boxed{P(A^c\cap B^c)=0.4}

c. DeMorgan's law can be used here too:

P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)

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\boxed{P(A^c\cup B^c)=0.9}

4 0
3 years ago
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