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3 years ago

A and B are vertical angles. If m_A = (3x - 30)° and m

Mathematics

1 answer:

Vilka [71]3 years ago

4 0

<h3>Answer: 21</h3>

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

Work Shown:

Vertical angles are always congruent.

This means they are equal in measure.

A = B

3x-30 = 2x-9

3x-2x = -9+30

x = 21

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Side note:

if x = 21, then,

A = 3*x-30 = 3*21-30 = 33 degrees

B = 2*x-9 = 2*21-9 = 33 degrees

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Find the exact values of the remaining five trigonometric functions of o satisfying the given conditions.

Lyrx [107]

First, tan(θ) = sin(θ) / cos(θ), so if cos(θ) = 3/5 > 0 and tan(θ) < 0, then it follows that sin(θ) < 0.

Recall the Pythagorean identity:

sin²(θ) + cos²(θ) = 1

Then

sin(θ) = -√(1 - cos²(θ)) = -4/5

and so

tan(θ) = (-4/5) / (3/5) = -4/3

The remaining trig ratios are just reciprocals of the ones found already:

sec(θ) = 1/cos(θ) = 5/3

csc(θ) = 1/sin(θ) = -5/4

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3 years ago

(-5) -2-8 + 1 1-5-2-8 helpp plisssssss

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Answer:

-14

Step-by-step explanation:

(-5)-2-8+1

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3 years ago

Read 2 more answers

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

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2 years ago

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The answer is C. Hope this helps

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3 years ago

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kondor19780726 [428]

Answer:

Step-by-step explanation:

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2 years ago