Mathematics, 05.10.2019 19:30 naomicervero

You're flying from joint base lewis-mcchord (jblm) to an undisclosed location 125 km south and 135 km east. mt. rainier is located 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?
convert hours into minutes in the final solution.

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Answer from: BM213

5.12 minutes ≈ 5 minutes 7.2 seconds

Step-by-step explanation:

The flight path of the airplane is along the line ...

y = -125/135x = -25/27x

The perpendicular line through Mt. Rainier's location is ...

y = 27/25(x -56) -40

In standard form, these two equations are ...

25x +27y = 027x -25y = 2512

The point of closest approach has the coordinates that are the solution to these two equations:

(x, y) = (33912/677, 31400/677)

The distance d from the origin to this point is given by the Pythagorean theorem (distance formula) as ...

d = 1256√(2/677) . . . . . km

This distance can be converted to time using the speed of the airplane:

$(1256\sqrt{\dfrac{2}{677}}\,km)\times \dfrac{60\,min}{800\,km}=94.2\sqrt{\dfrac{2}{677}}\,min\approx 5.120019\,min$

5.12 minutes after departing JBLM you will be closest to Mt. Rainier.

_____

The attached graph shows the geometry of the problem.

You're flying from joint base lewis-mcchord (jblm) to an undisclosed location 125 km south and 135 k

Answer from: blackordermember

The closest distance from a point to a line is its perpendicular distance to the line

The time it would take to be closest to Mt Rainier, is approximately 5.12 minutes

The reason why the value is correct is given as follows:

The given parameters are;

The direction of flight = 125km south and 135 km east

The location of Mt. Retainer = 56 km east and 40 km south of JBLM

The speed of flight = 800 km/hr

The slope of the travel path, m, is given as follows;

$m = -\dfrac{125}{135} = -\dfrac{25}{27}$

The equation of the path is therefore;

$y + 125 = -\dfrac{25}{27} \times (x - 135)$

$y = -\dfrac{25}{27} \times (x - 135) - 125$

The required slope for the perpendicular distance from the travel path to the Mt Rainier, m', is therefore;

$m' = -\dfrac{1}{m}$

Which gives;

$m' = -\dfrac{1}{-\dfrac{25}{27}} = \dfrac{27}{25}$

The equation of the line is therefore;

$y + 40 = \dfrac{27}{25} \times (x - 56)$

$y = \dfrac{27}{25} \times (x - 56) - 40$

The coordinates of the point where the two lines meet is therefore;

$\dfrac{27}{25} \times (x - 56) - 40 = -\dfrac{25}{27} \times (x - 135) - 125$

Solving with a graphing calculator gives;

x ≈ 50.09

y ≈ -46.38

The magnitude of the distance in km is, d = $\sqrt{(-46.38^2 + 50.09^2)} \approx 68.267$

The time it will take to be closest to Mt. Rainier, t, is given as follows;

$t \approx \dfrac{68.267 \, km}{800 \, km/hour} \approx 8.53 \times 10^{-2} \, hour$

The time it would take to get closest to Mt Rainier, t ≈ 8.53 × 10⁻² hours

$The \ time \ in \ minutes, \, t = 8.53 \times 10^{-2} \ hour\times \dfrac{60 \, min}{hour} \approx \underline {5.12 \ minutes}$

Time it would take to be closest to Mt Rainier in minutes, t ≈ 5.12 minutes

Learn more about distance from a line to a point here:

link

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

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

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is