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zaharov [31]
3 years ago
15

One month,ruby worked 6 hours more than Isaac, and Svetlana worked 4 times as many hours as ruby. Together they worked 126 hours

find the number of hours each person worked.
Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0
I + 6 = r

4r = s

i + r + s = 126

4i + 24 = 4r

s= 4r

i = r-6

4i+24=i+6
-3i=18

i = 6

Solve the rest
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In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + ta
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Answer:

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Step-by-step explanation:

The question is not properly formatted. However, the integral of \int {7 \sec(\theta) } \, d\theta is as follows:

<h3></h3>

\int {7 \sec(\theta) } \, d\theta

Remove constant 7 out of the integrand

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta

Multiply by 1

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta

Express 1 as: \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta

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\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta

Let

u = \sec(\theta) + \tan(\theta)

Differentiate

\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)

Make d\theta the subject

d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}

So, we have:

\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}

Cancel out \sec(\theta)\tan(\theta) + sec^2(\theta)

\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}

Integrate

\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c

Recall that: u = \sec(\theta) + \tan(\theta)

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