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Oduvanchick [21]
2 years ago
14

In the triangular prism shown below, which lines are parallel?

Mathematics
2 answers:
maw [93]2 years ago
6 0

Answer:

the 3rd one

Step-by-step explanation:

musickatia [10]2 years ago
3 0

Answer:

JM and KN

Step-by-step explanation:

one thing you can do to eliminate some answer choices is to make sure that the statement doesn't repeat a letter

the correct answer would be something like AB and BC, it would be like AB and CD

from there, you just have to look and pick your answer

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Step-by-step explanation:

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3 years ago
Determine the equation of the line that is perpendicular to the lines r(t)=(-2+3t,2t,3t)
Mnenie [13.5K]
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5) 
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t 
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t 
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-2 = t
For y sub in -8 
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(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t 
(-1 - 5)/4 = t
-1 = t
For y sub in -7 
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points: 
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int) 
Sub t = 2/3 into x = 5 + 4t 
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3) 
x = 23/3
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aleksley [76]
Answer: Quadratic

Solution
X=0
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6 0
3 years ago
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