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Alisiya [41]
2 years ago
11

Look at the two patterns below: Pattern A: Starts at 5 and follows the rule "add 6" Pattern B: Starts at 5 and follows the rule

"add 3" Select the statement that is true. (4 points) a The terms in Pattern A are 2 times the corresponding terms in Pattern B. B The terms in Pattern B are one third the corresponding terms in Pattern A. C The first five terms in Pattern A are 5, 11, 17, 23, and 29. D The first five terms in Pattern B are 0, 5, 8, 11, 14.
Mathematics
1 answer:
OleMash [197]2 years ago
8 0

Answer:

The first five terms in Pattern A are 5, 11, 17, 23, and 29

Step-by-step explanation:

Pattern A:

5 + 6a

Pattern B:

5 + 3a

Take a = 1 and compare ;

Pattern A:

5 + 6(1) = 11

Pattern B:

5 + 3(1) = 8

A ≠ 2B

B ≠ 1/3A

First 5 terms in pattern A :

Pattern A:

5

5 + 6(1) = 11

5 + 6(2)= 17

5 + 6(3) = 23

5 + 6(4) = 29

5 + 6(5) = 35

Pattern B cannot have a term 0 as it starts from 5

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

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Answer:

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