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sergij07 [2.7K]
3 years ago
10

Use the Pythagorean identity to do the following:

Mathematics
1 answer:
nlexa [21]3 years ago
7 0

Answer:  2 - 2*sin³(θ) - √1 -sin²(θ)

Step-by-step explanation:  In the expression

cos(theta)*sin2(theta) − cos(theta)

sin (2θ) = 2 sin(θ)*cos(θ)     ⇒   cos(θ)*2sin(θ)cos(θ) - cos(θ)

2cos²(θ)sin(θ) - cos(θ)           if we use cos²(θ) = 1-sin²(θ)

2 [ (1 - sin²(θ))*sin(θ)] - cos(θ)

2  - 2sin²(θ)sin(θ) - cos(θ)  ⇒  2-2sin³(θ)-cos(θ)   ;  cos(θ) = √1 -sin²(θ)

2 - 2*sin³(θ) - √1 -sin²(θ)

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2 years ago
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After graduating from college, Jane has two job offers to consider. Job A is compensated at $100,000 a year but
vaieri [72.5K]

Answer: (a)  Job A , by approximately $69, 482

\\ (b)  Job A , by approximately $6,867

 \\(c) Job B , by approximately $ 767,362

  \\ (d) Job A

Step-by-step explanation:

JOB B

\\The starting Salary is $ 10,000

Since there is an increment of 25% at the beginning of each new year. The breakdown of the increment is as follow:

\\First year : 125% of $ 10,000 = $12,500

\\Second year : 125% of $ 12,500 = $15,625

\\Third year : 125% of $15,625= $19,531.25

\\Fourth year: 125% of $19,531.25 = $24,414.06

\\Fifth year : 125% of $24,414.06 = $30,517.58

\\Sixth year: 125% of $30,517.58 = $38,147.00

\\Seventh year: 125% of $38,147.00 = $47,683.75

\\Eight year: 125% of $47,683.75 = $59,604.69

\\Ninth year: 125% of $59,604.69 = $74,505.86

\\Tenth year: 125% of $74,505.86 = $93,132.32

\\Following the same procedure:

\\Eleventh year = $116,415.40

\\Twelfth year = $145,519.25

\\Thirteenth year = $181,899.06

\\Fourteenth year = $227,373.82

\\Fifteenth year = $ 284,217.29

\\Sixteenth year = $355,271.61

\\Seventeenth year = $444,089.51

\\Eighteenth year = $555,111.89

\\Nineteenth year = $693,889.86

\\Twentieth year = $867,362.32

\\(a) Following the analysis above, at the beginning of the fifth year Job A will have a greater annual salary

\\Difference: Salary of Job A at the beginning of fifth year remains $ 100,000 while that of Job B resulted into $ 30,517.58, the difference implies

$100,000 - $ 30,517.58 = $69,482

\\(b) At the beginning of tenth year, Job A is still $100,000, Job B resulted into $93,132.32. Job A is still greater by approximately $6,867

\\(c) At the beginning of the twentieth year, the annual salary of A is still $ 100,000 while the annual salary of B is $ 867,362.32. Job B annual salary is greater by approximately $ 767,362

\\(d) If I were in Jane’s shoe I will take Job A and work for few years to gain more experience the look for a job that pays better. Waiting for many years in case of Job B is risky , market situation is uncertainty.

8 0
3 years ago
X - 2y = -3<br> 2x + y = -1
bezimeni [28]

Answer:

y = 1

x = -1

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x - 2y = -3, then x = 2y - 3

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substitute for x:

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4y - 6 + y = -1

5y = 5

y = 1

x = 2 - 3 = -1

3 0
2 years ago
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Paha777 [63]

Answer: Choice A)

H = \frac{SA}{2(L+W)} - \frac{LW}{L+W}\\\\

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

Work Shown:

SA = 2(LW + LH + HW)\\\\2(LW + LH + HW) = SA\\\\LW + LH + HW = \frac{SA}{2}\\\\LH + HW = \frac{SA}{2} - LW\\\\H(L + W) = \frac{SA}{2} - LW\\\\H = \frac{\frac{SA}{2} - LW}{(L + W)}\\\\H = \frac{SA}{2(L+W)} - \frac{LW}{L+W}\\\\

This is not the only way to solve for H.

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2 years ago
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