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3 years ago

I need to find the value of x given the answer in simplest radical form

Mathematics

1 answer:

Dominik [7]3 years ago

6 0

Answer:

knnnnndndnddndndn is a great place

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3 years ago

Read 2 more answers

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

Which of the following is an example of a perfect square trinomial?

Aliun [14]

Answer:

Step-by-step explanation:

A perfect square trinomial is always in the form

(a+b)(a+b)= $a^{2} +2ab+b^2$

ie. (x+4)(x+4)= $x^{2} +8x+16$

4 0

3 years ago

For my problem I must know the percent increase for;

PolarNik [594]

I think that u would have to take 14 from 13 and divide that by 2 (I THINK) i am not fully sure though

4 0

3 years ago

Can someone help with this I didn’t pay attention in math and I have homework ‍♀️

Lady bird [3.3K]

Answer:

Step-by-step explanation:

Even though it is absolute value, 17 is still the greatest answer. If a number is in absolute value it will never be negative, only positive values can come out of an absolute value.

An example for absolute value would be if you had to measure the amount of distance away from something, even if its behind you, it would still be a positive number of length away from you.

If you had the absolute value of -25, it would automatically be equal to 25.

So circling back to the original problem, the absolute value of 17 would be greater than 3, -4 and -8. If (hypothetically) it was the absolute value of -17 it would still be greater because it would equal 17.

I hope that made sense, im sorry that it's a bit lengthy. Have a good day :)

7 0

2 years ago