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Mashutka [201]
3 years ago
9

In numb3rs, they use math to solve crimes. One of the ways they use Math is to reveal patterns. What is the pattern in this sequ

ence? What would be the first and last numbers be? (6, 14, 30, 62, 126)
Mathematics
1 answer:
Mama L [17]3 years ago
5 0

Step-by-step explanation:

I think the first number is 2

Then 2 x 2 = 4 so there is a difference of 4 number

Then in between 6 and 14 there is a difference of 8 so 4 x 2 = 8

Then 16 so 8 x 2 = 16

So after 126 the number is 254

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I believe it is 2 but I am not sure
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What are the terms in the expression 2d + 7 + 5f?
expeople1 [14]

Answer: 2d, 7, and 5f

Step-by-step explanation: Each part of the equation that is separated by a + or - is considered a "term."

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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
Maurinko [17]

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

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)

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

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