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Alchen [17]
3 years ago
10

(1 point) find the point pp where the line x=1+t,y=2t,z=−3tx=1+t,y=2t,z=−3t intersects the plane x+y−z=1.

Mathematics
1 answer:
emmasim [6.3K]3 years ago
5 0
Sub in values for x,y,z into plane equation.

(1+t) + 2t - (-3t) = 1

Solve for 't'

6t + 1 = 1
t = 0

Sub back in t=0 into line to find point.
x = 1+0 = 1
y = 2(0) = 0
z = -3(0) = 0

Answer :  The point is (1,0,0)
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Answer:

62°

Step-by-step explanation:

The angle R inscribes the arc FQ, so using the property of inscribed angles in a circle, we have that:

m∠R = mFQ / 2

The arc FQ is the sum of the arcs FP and PQ, so we have:

mFQ = mFP + mPQ = 11x + 7 + 60 = 11x + 67

Now, with the first equation, we have:

12x + 1 = (11x + 67) / 2

24x + 2 = 11x + 67

13x = 65

x = 5°

So we have that mFP = 11x + 7 = 55 + 7 = 62°

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2 years ago
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Step-by-step explanation:

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3 years ago
The variables x, y, and z vary jointly.
serious [3.7K]

Answer:

k = 18

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Step-by-step explanation:

x = k * y * z

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find z when x = -3 and y = 4

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3 years ago
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Answer:

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7 0
3 years ago
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