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agasfer [191]
4 years ago
7

A rectangle’s area and perimeter have the same numerical value. The length of the rectangle is 10 feet.

Mathematics
1 answer:
KiRa [710]4 years ago
3 0

Answer:

The width of the rectangle is 2.5 feet

Step-by-step explanation:

Here, we want to calculate the width of the rectangle given its length

Let the area = perimeter = x

Mathematically;

area of rectangle = L * B = 10B

The perimeter of the rectangle = 2(L + B)

Also;

x = 10B

Let’s equate this to the perimeter

x = 10B = 2(L + B)

10B = 2(10 + B)

10B = 20 + 2B

10B -2B = 20

8B = 20

B = 20/8

B = 5/2

B = 2.5

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Answer:

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

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

Expand

\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

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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)

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

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