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ddd [48]
2 years ago
5

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

4(θ) dθ
Mathematics
1 answer:
ozzi2 years ago
7 0

I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta  \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt

by using the integration formula

we get,

\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c

now put the value of t=\sin\theta in the above equation

we get,

\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c

hence proved

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Answer:

Step-by-step explanation:

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P(unlocked homes) = P(U)

P(unlocked homes) = 40%

P(unlocked homes) = 0.40

Also, let us represent the value of the locked room with G.

P(locked homes) = P(G)

P(locked homes) = (1 - 0.40)

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Let the probability of selecting a correct key be P(S)

It implies that for the agent to use 3 keys, we have a combination of ^8C_3 possible ways for the set of keys.

Now; since only one will open the house, then:

P(select correct key) = P(S)

P(S) = \dfrac{ (^1_1) (^7_2)   }{   ^8_3 }

P(S) = \dfrac{21}{56}

P(S) = 0.375

Finally, for the real estate agent to have access to specific homes supposing the agent select three master keys at random prior to the time he left his office, Then:

P(F ∪ (G∩S) = P(F) + P(G∩S)

P(F ∪ (G∩S) = P(F) + P(G) × P(S)

P(F ∪ (G∩S) = 0.40 + (0.60×0.375)

P(F ∪ (G∩S) = 0.40 + 0.225

P(F ∪ (G∩S) =0.625

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Tyler concludes that 5x² will always have a larger output for the same value of x.

<u>Look at the graph below and the table given</u>

Take a random value: x = 0

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Here, 1 > 0, making 2^x > 5x²

Hence, 2^x is greater than 5x² at this point. making Tyler's point not applicable.

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