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lana [24]
3 years ago
8

What is the exact value of cos (67.5° )?

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

0.38268343

Step-by-step explanation:

cos( 67.5 )

Rewrite  

67.5

as an angle where the values of the six trigonometric functions are known divided by  2 . cos ( 135\ 2)

Apply the cosine half-angle identity.

±

√

1

+

cos

(

135

)

2

 

Change the  

±

to  

+

because cosine is positive in the first quadrant.

√

1

+

cos

(

135

)

2

Simplify  

√

1

+

cos

(

135

)

2

√

2

−

√

2

2

The result can be shown in multiple forms.

Exact Form:

√

2

−

√

2

2

Decimal Form:

0.38268343

…

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What are the values of the trigonometric ratios for this triangle?
nika2105 [10]

Answer:

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6 0
2 years ago
Joel can either work 2 days for 11 hours each day or 4 days for 6 hours each day. He takes the bus for $2.25 each way. On days h
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7 0
3 years ago
Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th
DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that A \cap B = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

A + (A \cap B) = 0.40

A + 0.17 = 0.40

A = 0.23

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

B + (A \cap B) = 0.35

B + 0.17 = 0.35

B = 0.18

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0
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Assoli18 [71]

Answer:

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Step-by-step explanation:

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3 years ago
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