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Alchen [17]
1 year ago
9

Point X is (3, -6). Wgich point is 10 units away from Point X

Mathematics
1 answer:
vampirchik [111]1 year ago
6 0

If we find the point X on the plane we can see the following:

Notice that the point D and the point X are 10 units apart with respect the x-axis, therefore, the point that is 10 units away from X is point D

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How do you write 3 2/4 as a fraction greater t hff an 1
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(5x^4-2x^3-7x^2-39)÷(x^2+2x-4)
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4 0
2 years ago
Find an equation of the plane consisting of all points that are equidistant from (-3, 5, -4) and (-5, 0, 4), and having -2 as th
Firdavs [7]

Answer:

-2x-5y+8z+4.5=0

Step-by-step explanation:

Let (x,y,z) be the coordinates of the point lying on the needed plane. This point is equidistant from the points (-3, 5, -4) and (-5, 0, 4), so

d_1=\sqrt{(x-(-3))^2+(y-5)^2+(z-(-4))^2}=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}\\ \\d_2=\sqrt{(x-(-5))^2+(y-0)^2+(z-4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\d_1=d_2\Rightarrow \sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2\\ \\x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16\\ \\-4x-10y+16z+9=0\\ \\-2x-5y+8z+4.5=0

5 0
3 years ago
Read 2 more answers
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