All categories

3 years ago

Which of the following are concidered skew lines?

Mathematics

1 answer:

Fiesta28 [93]3 years ago

5 0

I think the answer is AE &GH

You might be interested in

Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximat

photoshop1234 [79]

Answer:

With n = 25, $\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549$

With n = 50, $\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594$

b. $\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943$

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

$\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 25.

Therefore,

$\Delta{x}=\frac{1-0}{25}=\frac{1}{25}$

We need to divide the interval [0,1] into n = 25 sub-intervals of length $\Delta{x}=\frac{1}{25}$ , with the following endpoints:

$a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

$2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579$

...

$2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701$

$f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549$

To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 50 with the trapezoid rule you must:

We have that a = 0, b = 1, n = 50.

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594$

b. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using 2n with the Simpson's rule you must:

The Simpson's rule states that

$\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 50

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the Simpson's rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943$

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by $|A-B|$

The absolute error in the trapezoid rule is

The calculated value is

$\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225$

and our estimate is 67.1519320308594

Thus, the absolute error is given by

$|67.0714655821225-67.1519320308594|=0.08047$

The absolute error in the Simpson's rule is

$|67.0714655821225-67.0715427161943|=0.00008$

6 0

3 years ago

What does 3w =30 mean a. 6 b.9 c.10 d.12

PolarNik [594]

Answer:

w = 10

Step-by-step explanation:

3w = 30

Divide each side by 3

3w/3 = 30/3

w = 10

8 0

3 years ago

Read 2 more answers

Draw a 270 degrees clockwise

Brilliant_brown [7]

You can't quite do that

6 0

3 years ago

A sport store ordered 22 boxes of golf balls. Each box has 47 golf balls. Which is the best estimate of the total number of golf

RoseWind [281]

About 1,000! Hope this helps

5 0

3 years ago

Read 2 more answers

Roberta makes $9 an hour plus a 10.5% commission selling jewelry an hour shift in which she sells $880 worth of jewelry. How muc

timama [110]

Answer:

Roberta earns $164.4 on an 8 hr shift.

Step-by-step explanation:

In one hour Roberta makes $9

So if she works for 8 hours she will be making

=> ( 9 X 8)$

=>72$

Also she earns a commission of 10.5% on selling a jewellery worth $880

So 10.5% of 880 is

=> $\frac{10.5}{100}\times 880$

=> $0.105\times 880$

=> 92.4

Hence Roberta will be earning a total amount of

(72+ 92.4)

=>$164.4

3 0

3 years ago