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Several hours after departure the two ships described to the right are 50 miles apart. If the ship traveling south traveled 10 miles farther that the other how many miles did they each travel.

Sagot :

MrRoyal

Answer:

[tex]Ship\ A = 40\ miles[/tex]

[tex]Ship\ B = 30\ miles[/tex]

Step-by-step explanation:

Given

[tex]d = 50[/tex] --- distance apart

[tex]Ship\ B = x[/tex]

[tex]Ship\ A = x + 10[/tex]

The question can be represented with the attached image.

Required

Solve for x

Using Pythagoras theorem, we have:

[tex]x^2 + (x + 10)^2 = 50^2[/tex]

Open bracket

[tex]x^2 + x^2 + 20x + 100 = 2500[/tex]

Collect like terms

[tex]x^2 + x^2 + 20x + 100 - 2500 = 0[/tex]

[tex]2x^2 + 20x -2400 = 0[/tex]

Divide through by 2

[tex]x^2 + 10x -1200 = 0[/tex]

Expand

[tex]x^2 + 40x - 30x -1200 = 0[/tex]

Factorize

[tex]x(x + 40) - 30(x +40) = 0[/tex]

Factor out x + 40

[tex](x - 30)(x +40) = 0[/tex]

Split and solve for x

[tex]x - 30 = 0\ or\ x + 40 = 0[/tex]

[tex]x = 30\ or\ x = -40[/tex]

Distance can not be negative

So:

[tex]x = 30[/tex]

Recall that:

[tex]Ship\ B = x[/tex]

[tex]Ship\ A = x + 10[/tex]

This implies that:

[tex]Ship\ B = 30\ miles[/tex]

[tex]Ship\ A = 30 + 10\ miles[/tex]

[tex]Ship\ A = 40\ miles[/tex]

View image MrRoyal
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