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3 years ago

A car travels 250 km east, then turns 25° south of east and goes 115 km. What is the resultant distance traveled?

Mathematics

1 answer:

Firlakuza [10]3 years ago

4 0

Answer:

The distance traveled is 357.54km.

Step-by-step explanation:

The situation is illustrated in the figure attached.

From the law of cosines the distance $d$ is given by

$d^2 = (250km)^2+(115km)^2-2(250km)(115km)cos(155^o)$

$d^2 = 127837.698 km^2$

$\boxed{d = 357.54km}$

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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

Function A and Function B are linear functions. Compare the two functions and choose all that are correct.

Olegator [25]

Answer:

2 .The slope of Function A is less than the slope of Function B

Step-by-step explanation:

A graph of Function A shows it has a y-intercept of 4, the same as that of Function B. (Statements 3 and 4 are not correct.)

The slope of Function A is 2, which is less than the slope of 3 that Function B has. (Statement 2 is correct; statement 1 is not.)

_____

<em>More detailed working</em>

The slope of Function A can be figured easily between the points with x-values that differ by 1:

m = (y3 -y2)/(x3 -x2) = (24-22)/(10-9) = 2/1 = 2 . . . . . Fun A has slope of 2.

The slope of Function B is the coefficient of x in the equation: 3.

The y-intercept of Function A can be found starting with point-slope form:

y -22 = 2(x -9)

y = 2x -18 +22

y = 2x +4 . . . . . . . slope-intercept form

The intercept of +4 is the same as that of Function B.

5 0

2 years ago

Read 2 more answers

What is 4 - ( -h ) = 68?

Kryger [21]

4 + h = 68

h = 68 - 4

h = 64

hope this helps

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2 years ago

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twenty students in A and 20 students in class B were asked how many hours they took to prepare for an exam.The data sets represe

SSSSS [86.1K]

Answer:

is there a graph i can read?

Step-by-step explanation:

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2 years ago

Joe's annual income has been increasing each year by the same dollar amount. The first year his income was $17 comma 90017,900

Vedmedyk [2.9K]

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points $(x_1,y_1)$ and $(x_2,y_2)$ , then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line is

$y-17900=\frac{20300-17900}{4-1}(x-1)$

$y-17900=\frac{2400}{3}(x-1)$

$y-17900=800(x-1)$

$y-17900=800x-800$

Add 17900 on both sides.

$y=800x-800+17900$

$y=800x+17100$

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

$30700=800x+17100$

Subtract 17100 from both sides.

$30700-17100=800x$

$13600=800x$

Divide both sides by 800.

$\frac{13600}{800}=x$

$17=x$

Therefore, in 17th year his income was $30,700.

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3 years ago