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3 years ago

Answer it answer it answer it

Mathematics

2 answers:

Troyanec [42]3 years ago

8 0

Answer:

22.5

Step-by-step explanation:

cause

Lana71 [14]3 years ago

6 0

Answer:

It is 22.5

Step-by-step explanation:

add up all they shaded parts:

22.5 x 4 = 90 degrees

as each unshaded part is three times more then you do 22.5 x 3 for one part

22.5 x 3 = 67.5 degrees

then times by four because there is four unshaded sections.

67.5 x 4 = 270 degrees

270 + 90 = 360 degrees and there is 360 degrees in a circle so 22.5 degrees is the correct answer.

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20 points One diagonal of a rhombus has endpoints (-10, 1) and (2, 9).

Makovka662 [10]

Check the picture below.

so as you already know, a rhombus is a parallelogram whose sides are equal, so the distance from say (-10, 1) to either endpoint of the other diagonal must be the same.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{1})\qquad \underline{(\stackrel{x_2}{-2}~,~\stackrel{y_2}{2})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-2-(-10)]^2+[2-1]^2}\implies d=\sqrt{(-2+10)^2+(2-1)^2} \\\\\\ d=\sqrt{64+1}\implies \boxed{d=\sqrt{65}} \\\\[-0.35em] ~\dotfill$

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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

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