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3 years ago

Which value is a solution for the equation tan x/2 = -1?

Mathematics

2 answers:

Luba_88 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

What we have to do is to find the arc tan

x/2 = arc tan (-1)

tan is negative on the 4th and second quadrant

arc tan (-1) = 135 or 315

pi is 180 degrees

so we have;

135 is 3 pi/4

multiplied by the 2

= 6 pi/4 which is 3 pi/2

315 is 7 pi/4 which is same as 7 pi/2 when we multiply by 2

zysi [14]3 years ago

3 0

Answer:

the answer is d

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Nezavi [6.7K]

Total number of pieces of paper = 11
Number of pieces with a vowel = 4

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3 years ago

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A bag of rice weighs 3.18 pounds. Find its mass in kilograms. (1 kg ≈ 2.2 lb)

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Answer:

1.40614

Step-by-step explanation:

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3 years ago

If the vertex is (2, -8), what is the axis of symmetry? *

frozen [14]

Answer:

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3 years ago

You are making an audience selection. The first selection narrows your audience of 240,000 by 25% and the next selection narrows

vlada-n [284]

Answer:

the answer is c. 108,000

Step-by-step explanation:

240,000×25%=60,000

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8 0

3 years ago

An electronics hobbyist has three electronic parts cabinets with two drawers each.

Andrews [41]

Answer:

a) 0.5 = 50% probability that an NPN transistor will be selected.

b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types

c) 66.67% probability that it comes from the cabinet that contains only NPN transistors

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

a) What is the probability that an NPN transistor will be selected?

1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.

$p = \frac{1}{3}*1 + \frac{1}{3}*0 + \frac{1}{3}*0.5 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = 0.5$

0.5 = 50% probability that an NPN transistor will be selected.

b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?

Here we use the conditional probability formula.

Event A: NPN transistor

Event B: From the third cabinet.

50% probability that an NPN transistor will be selected, so $P(A) = 0.5$ .

1/6 probability that it is from the third cabinet and NPN, so $P(A \cap B) = \frac{1}{6}$

The desired probability is:

$P(B|A) = \frac{\frac{1}{6}}{0.5} = 0.3333$

0.3333 = 33.33% probability that it came from the cabinet that contains both types.

c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?

Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So

x + 33.33 = 100

x = 66.67

66.67% probability that it comes from the cabinet that contains only NPN transistors

6 0

3 years ago