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3 years ago

What is the best answer for sin(\dfrac{2}{x})=0

Mathematics

1 answer:

nexus9112 [7]3 years ago

3 0

The sine of an angle is zero if and only if the angle is zero, with a periodicity of $\pi$

In other words, you have

$\sin(\alpha)=0\iff\alpha=0,\ \pm\pi,\ \pm 2\pi\ldots = k\pi,\ k\in\mathbb{Z}$

So, we have

$\sin\left(\dfrac{2}{x}\right)=0 \iff \dfrac{2}{x}=k\pi \iff x =\dfrac{2}{k\pi}$

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The diagram below shows a cross-section of a metal pipe, that has a length of 50 cm.

nikitadnepr [17]

Answer:

100

Step-by-step

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Sam bought 9 ounces of cheese for $3.24. Joe bought 7 ounces of cheese for $3.22. Who got the best deal?

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Describe a method you can use to find the area of the following shape. Provide specific details in your answer, including coordi

insens350 [35]

The area of figure ABCDEF can be computed as the sum of the areas of trapezoid ACDF and triangle ABC, less the area of trangle DEF.
trapezoid ACDF area = (1/2)(AC +DF)·(CD) = (1/2)(8+5)(6) = 39
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2 years ago

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2 years ago

Read 2 more answers

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

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