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allochka39001 [22]
3 years ago
10

Find the points of intersection of the line x = 3 + 2t, y = 5 + 8t, z = −6 + t, that is, l(t) = (3 + 2t, 5 + 8t, −6 + t), with t

he coordinate planes. xy plane (x, y, z) = xz plane (x, y, z) = yz plane (x, y, z) =

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0
See, please, suggested decision:
When the line intersects the xy-plane, then parameter z=0, when does the xz - y=0 and when does yz, x=0. All the details are in attachment.
If it's possible, check arithmetic part.

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The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room
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I'm going to assume that the room is a rectangle.

The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle. 

You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
A = lw\\
600 = (x+5)(x+4)\\
600 =  x^{2} + 9x + 20


Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
600 = x^{2} + 9x + 20\\
x^{2} + 9x - 580 = 0\\
(x + 29)(x - 20) = 0\\
x + 29 = 0, \:\: x - 20 = 0\\
x = -29, x = 20

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.

The dimensions of your room are 25ft (length) by 24ft (width).
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