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DENIUS [597]
3 years ago
7

A jet plane traveling at 550 mph overtakes a propeller plane traveling at 150 mph that had a 3​-hour head start. How far from th

e starting point are the​ planes?
Mathematics
1 answer:
Otrada [13]3 years ago
8 0

The planes are 618.75 miles from starting point

<em><u>Solution:</u></em>

Let t is the travel time of the jet

Then  (t + 3) is the travel time of the propellar

When the jet overtakes the propellar, they will traveled the same distance

<em><u>Distance is given as:</u></em>

Distance = speed \times time

<h3><u>Distance equation for Jet</u></h3>

A jet plane traveling at 550 mph

Distance\ by\ jet = 550 \times t\\\\Distance\ by\ jet = 550t

<h3><u>Distance equation for propellar:</u></h3>

propeller plane traveling at 150 mph

Distance\ by\ propellar = 150 \times (t+3)

Equate both distance

550t = 150(t+3)\\\\550t = 150t + 450\\\\550t - 150t = 450\\\\400t = 450\\\\t = \frac{450}{400}\\\\t = 1.125

<h3>Find the distance from the starting point</h3>

Distance = 550 \times t = 550 \times 1.125\\\\Distance = 618.75

Thus they are 618.75 miles from starting point

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The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

The given parameters are:

\mathbf{Height = 20}

\mathbf{Length = 28}

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

\mathbf{Mid = \frac 12 \times Length}

\mathbf{Mid = \frac 12 \times 28}

\mathbf{Mid = 14}

So, the vertex of the parabola is:

\mathbf{Vertex = (Mid,Height)}

Express properly as:

\mathbf{(h,k) = (14,20)}

A point on the graph would be:

\mathbf{(x,y) = (28,0)}

The equation of a parabola is calculated using:

\mathbf{y = a(x - h)^2 + k}

Substitute \mathbf{(h,k) = (14,20)} in \mathbf{y = a(x - h)^2 + k}

\mathbf{y = a(x - 14)^2 + 20}

Substitute \mathbf{(x,y) = (28,0)} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{0 = a(28 - 14)^2 + 20}

\mathbf{0 = a(14)^2 + 20}

Collect like terms

\mathbf{a(14)^2 =- 20}

Solve for a

\mathbf{a =- \frac{20}{14^2}}

\mathbf{a =- \frac{20}{196}}

Simplify

\mathbf{a =- \frac{5}{49}}

Substitute \mathbf{a =- \frac{5}{49}} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

Hence, the equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

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2 years ago
Hellppp me this is hard​
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Answer:

f

Step-by-step explanation:

its is that because it would mean he would be going 65 whuch its asking for a general speed.

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