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vichka [17]
2 years ago
7

An airplane is flying at a speed of 500 mi/h at an altitude of one mile. The plane passes directly above a radar station at time

t = 0.
(a) Express the distance s (in miles) between the plane and the radar station as a function of the horizontal distance d (in miles) that the plane has flown.
s(d) =
(b) Express d as a function of the time t (in hours) that the plane has flown.
d(t) =
(c) Use composition to express s as a function of t.
s(t) =
Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
6 0

Answer:

a)s=\sqrt{1+d^2}

b)d(t)=500t

c)s(t) =\sqrt{1+250000t^2}

Step-by-step explanation:

d = Horizontal distance

s = the distance between the plane and the radar station

The horizontal distance (d), the one mile altitude, and s form a right triangle.

So, use Pythagoras theorem

Hypotenuse^2=Perpendicular^2+Base^2

s^2=1^2+d^2

a) s=\sqrt{1+d^2}

(b) Express d as a function of the time t (in hours) that the plane has flown.

distance = speed \times time

d(t)=500t

(c) Use composition to express s as a function of t.

s(t) =\sqrt{1+d^2}

using b

s(t) =\sqrt{1+(500t)^2}\\s(t) =\sqrt{1+250000t^2}

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Find an equation of the line containing the centers of the two circles whose equations are given below.
Anna35 [415]

Answer:

<h2><em>3y+x = -5</em></h2>

Step-by-step explanation:

The general equation of a circle is expressed as x²+y²+2gx+2fy+c = 0 with centre at C (-g, -f).

Given the equation of the circles x²+y²−2x+4y+1  =0  and x²+y²+4x+2y+4  =0, to  get the centre of both circles,<em> we will compare both equations with the general form of the equation above as shown;</em>

For the circle with equation x²+y²−2x+4y+1  =0:

2gx = -2x

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Divide both sides by 2:

2g/2 = -2/2

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Also, 2fy = 4y

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The centre of the circle is (-(-1), -2) = (1, -2)

For the circle with equation x²+y²+4x+2y+4  =0:

2gx = 4x

2g = 4

Divide both sides by 2:

2g/2 = 4/2

g = 2

Also, 2fy = 2y

2f = 2

f = 1

The centre of the circle is (-2, -1)

Next is to find the equation of a line containing the two centres (1, -2) and (-2.-1).

The standard equation of a line is expressed as y = mx+c where;

m is the slope

c is the intercept

Slope m = Δy/Δx = y₂-y₁/x₂-x₁

from both centres, x₁= 1, y₁= -2, x₂ = -2 and y₂ = -1

m = -1-(-2)/-2-1

m = -1+2/-3

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The slope of the line is -1/3

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Substituting the point (-2, -1) and slope of -1/3 into the equation y = mx+c

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y = -1/3 x + (-5/3)

y = -x/3-5/3

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<em>Hence the equation of the line containing the centers of the two circles is 3y+x = -5</em>

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