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3 years ago

ABCD is a quadrilateral inscribed in a circle, as shown below:

Mathematics

1 answer:

german3 years ago

3 0

Answer:

A. (x + 16) + (6x − 4) = 180

Step-by-step explanation:

Inscribed quadrilateral has opposite angles supplementary.

m∠A + m∠C = 180°

m∠B + m∠D = 180°.

<u>Since</u>

m∠A = x + 16,
m∠B = x,
m∠C = 6x - 4,
m∠D = 2x + 16

we can use either pair of angle measures to work out the value of x and then find the value of each angle.

<u>We can verify the first option is the only correct one.</u>

m∠A + m∠C = 180°
(x + 16) + (6x − 4) = 180

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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

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

4 0

3 years ago

This year Maya watched 60 movies. She thought that 33 of them were very good. Of the movies she watched, what percentage did she

Deffense [45]

Answer:

55%

Step-by-step explanation:

Divide 33 by 60 each percent is 1 2/3 so 30 is 50% and 3 is 5 50%+5%=55%

6 0

1 year ago

An insect crawls 30 inches in 16.3 minutes. If

omeli [17]

Answer:

Step-by-step explanation:

Convert 6 hours to minutes so that 6 hours and 16. 3 minutes can be in the same units.

Since 1 hour = 60 minutes, then 6 hours = 6 x 60 minutes = 360 minutes

Since the insect has more minutes to crawl (360 minutes), then we expect more inches to be covered. Therefore, the ratio is 360 minutes to 16.3 minutes

= 360 minutes : 16.3 minutes

= 360 : 16.3

Now express the ratio as fraction and relate it with 30 inches

= 360/16.3 x 30/1

= 10800/16.3

= 662 .576687 and this is approximately 650 inches

Answer = 650 = (C)

3 0

2 years ago

2. Round the following numbers to the nearest 10 thousand:

Vlad [161]

The values of the given numbers when it is rounded up to the nearest 10 thousands are:

990000
150000

<h3>What is rounding up in mathematics?</h3>

Rounding up can be described as the process that is been used in the mathematics which is been used in the estimation of a particular number in a context.

It should be noted that in rounding the a number up, it is required to look at the next digit at the right hand of the given figures in a case whereby the digit is less than 5,the digit can be rounded down, but in the case whereby the digit is more that 5 then it can be rounded up .

From the given values, we are given the 990,201 and 159,994 and if this were to rounded up to the nearest 10 thousand then we will start from the right hand sides and round down the values less than 5 and round up the values that is more that 5. and their values will be 990000

and 150000.