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Sonbull [250]
2 years ago
11

Find the point(s) of intersection (if any) of the plane and the line. Also, determine whether the line lies in the plane. 2x - 2

y + z = 12, x - 1/2 = y + (3/2)/-1 = (z + 1) / 2
Mathematics
1 answer:
s2008m [1.1K]2 years ago
7 0

Answer:

the intersection is the point P=(x,y,z)=(8,9,14) and the line does not lie in the plane

Step-by-step explanation:

from the equation of the line

x - 1/2 = y + (3/2)/-1 = (z + 1) / 2  = t (parameter)

then the parametric equation of the line is

x= 1/2 +t

y =  (3/2) + t

z = (-1) + 2*t

therefore from the equation of the plane

2x - 2y + z = 12

2*(1/2 +t) - 2[(3/2) + t]  + [(-1) + 2*t]  = 12

1+2*t - 3 -2*t -1 + 2*t = 12

-3 + 2*t = 12

t= 15/2

therefore there is only one intersection of the line with the plane ( then the line does not lie in the plane , since there would be infinite intersection points). The intersection is

x= 1/2 +t = 1/2 +15/2 = 8

y =  (3/2) + t = (3/2) + 15/2 = 9

z = (-1) + 2*t =  (-1) + 2*15/2  = 14

thus the intersection point P=(x,y,z)=(8,9,14)

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avanturin [10]

Answer: 2x ^4 +7x−4x ^2 −4

Step-by-step explanation:

2x ​^4 −x*3−4x ^​2 +10x−4

2x ^​4 −3x−4x ^​2 +10x−4

2x ^​4 +(−3x+10x)−4x ^2 ​​−4

2x ^4 +7x−4x ^2 −4

7 0
3 years ago
Amanda's book bag is 0.10. Some of Amanda's pens are removed and replaced with pink pens. Her book bag still contains a total of
Marta_Voda [28]

Answer:

The number of pink pens can be 19 .

Step-by-step explanation:

Total Number of pens = 20

Let x be the number of pink pens

We are supposed to find If the probability of choosing a pink pen is close to 1, how many pink pens would you expect to see in Amanda's bag

Probability of choosing a pink pen = \frac{\text{Number of pink pens}}{\text{Total Number of pens}}

Probability of choosing a pink pen =\frac{x}{20}

ATQ

\frac{x}{20}≈1

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So, the number of pink pens can be 19 .

6 0
3 years ago
How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?
irga5000 [103]
The equivalence n \equiv 5 \pmod{12}

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5


for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)


we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009


The last k which produces n<5000 is 416


For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.


Answer: 417

6 0
3 years ago
Desperate<br> please Hurry<br> Question Down Below
dsp73

Answer:

-1 1/3

Step-by-step explanation:

4-8=-4

-4/3

-1 1/3

Hope this has helped. have a good day.

5 0
2 years ago
Read 2 more answers
Use a scientific calculator to find the logarithm for each number rounded ro four decimal places. Then state characteristic and
anzhelika [568]
Using a scientific calculator:
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Rounded to four decimal places:
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6 0
3 years ago
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