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Phantasy [73]
3 years ago
5

The hypotenuse of a 30°-60°-90° triangle measures 10 inches. Which could be the length of a leg of the triangle?​

Mathematics
1 answer:
diamong [38]3 years ago
4 0

Answer:

6 and 8

Step-by-step explanation:

The hypotenuse of a 30°-60°-90° triangle measures 10 inches.

To find the length of one of the legs, we use the idea of Pythagorean Triples.

Pythagorean Triples are any set of three numbers that satisfies the Pythagorean Theorem. Some common examples are:

  • 3, 4 and 5
  • 5, 12 and 13
  • 9, 40 and 41

Note that in a Pythagorean Triple,

  • The longest length is always the Hypotenuse.
  • New Triples can be formed from product of existing triples.

In our given triangle, the Hypotenuse=10 Inches

Consider the Pythagorean Triple 3,4, and 5

  • 5 is the Hypotenuse
  • Multiply the Triples by 2, we obtain:
  • 6, 8 and 10 (in which 10 is the hypotenuse)

Therefore, 6 and 8 could be the length of a leg of the 30°-60°-90° triangle.

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Y_Kistochka [10]

Answer:

1 6/9

Step-by-step explanation:

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2 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
2 years ago
Which of the following is an irrational number?
Sladkaya [172]

Answer:

1.333...

Step-by-step explanation:

because irrational numbers can be repeating decimals, numbers that aren't perfect squares,and numbers that can not be in fraction form.

4 0
3 years ago
Given f(x) = 9x - 10, find f(8) *
Rama09 [41]

Answer:

\huge\boxed{f(8) = 62}

Step-by-step explanation:

\sf f(x) = 9x-10\\\\Put\ x = 8\\\\f(8) = 9(8) -10\\\\f(8) = 72-10\\\\f(8) = 62\\\\\rule[225]{225}{2}

Hope this helped!

<h2>~AnonymousHelper1807</h2>
3 0
3 years ago
Read 2 more answers
If these two shapes are similar, what is the measure of the missing length s?
kodGreya [7K]
I think it’s 5 but i’m not sure lol
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2 years ago
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