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3 years ago

Solve for theta Please answer both parts

Mathematics

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

a) tanθ = x/3

θ = arctan(x/3)

:::::

b) cosθ = 9/x

θ = arccos(9/x)

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HELP PLSSSSSSSSSSSSSSS ASAP

makvit [3.9K]

Answer:

Step-by-step explanation:

Sydney = x

Devaughn = 2x

x + 2x = 78

3x = 78

x = 78 / 3

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6 0

2 years ago

Read 2 more answers

The sum of two numbers is 54. The large number is 18 more than the sampler number. What are the numbers ?

Karolina [17]

Answer: a=36; b=18

Step-by-step explanation:

1. You know that the sum of two numbers is 54, then, you have:

a+b=54

2. According to the problem, the larger number is 18 more than the sampler number, this can be expressed as following:

a=b+18

3. Then, you must substitute the second equation into the first equation and solve for b:

b+18+b=54

b=18

4. Then the value of a is:

a+18=55

a=36

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3 years ago

- m + 62m2 + 1200<br> simplified

Gekata [30.6K]

3(41x+400) simplified = 123x+1200

3 0

3 years ago

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a gallery owner purchased a very old painting for $3,000. the painting sells at a 325% increase in price what is the retail pric

nata0808 [166]

300%= 9,000
20%= 60
5%= 15

9,000+60+15= 9,075
answer: £9,075

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3 years ago

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The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes

Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(completing the exam in one hour or less)

P(x < 60)

$P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 60) =0.023= 2.3\%$

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

$P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%$

c) P(completing the exam in more than 90 minutes)

P(x > 90)

$P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

$\dfrac{15.87}{100}\times 60 = 9.52\approx 10$

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0

3 years ago