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White raven [17]

3 years ago

10

You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 201 km south and 194 km east. Mt. Rainier is locat

ed approximately 56 km east and 40 km south of JBLM. If you are flying at a constant speed of 800 km/hr, how long after you depart JBLM will you be the closest to Mt. Rainier?

1 answer:

elixir [45]3 years ago

6 0

Answer:

The answer is "0.0846276476".

Step-by-step explanation:

Let all the origin(0,0) of JBLM be

Let the y-axis be north along with the vector j unit.

But along the + ve x-axis is east, and all along with the vector unit i.

And at (194,-201) that undisclosed position is

Mt. Ris (56,-40) at

Let the moment it gets close to the Mt. rainier bet.

Oh, then,

If we know, the parallel to the direction from the point was its nearest one to a path to a point,

Calculating slope:

$\to m=\frac{y_2-y_1}{x_2-x_1}\\$

points: (0,0) and (194,-201)

$\to m=\frac{-201 -0 }{194-0}\\\\\to m=\frac{-201}{194}\\\\\to m=-1.03$

equation of the line:

$\to y= mx+c\\\\\to y= -1.03\ x+c$

when the slope is perpendicular= $-\frac{1}{m}$

$= - \frac{1}{ -1.03}\\\\= \frac{1}{ 1.03}\\\\= 0.97$

perpendicular equation:

$\to y-(-40)=0.97 \times (x-56) \\\\\to y+40=0.97 \times (x-56) \\\\$

Going to solve both of the equations to have the intersection point,

We get to the intersect level in order to be at

(47.16,-48.5748)

so the distance from origin:

$= \sqrt{(47.16^2+48.5748^2)}\\\\= \sqrt{2224.0656 + 2359.5112}\\\\=\sqrt{4583.5768}\\\\=67.7021181$

$\to time =\frac{distance }{speed}$

$= \frac{67.7021181}{800}\\\\= 0.0846276476$

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Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

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Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

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Step-by-step explanation:

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Read 2 more answers

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4 years ago