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S_A_V [24]
3 years ago
5

Find the equation of a plane that is perpendicular to the vector −4i⃗ −4j⃗ −k⃗ and passing through the point (−2,−5,5)

Mathematics
1 answer:
vredina [299]3 years ago
5 0

hello ....

the equation of a plane that is ;  ax+by+cz +d =0

the vector perpendicular to this plane is : V(a,b,c)

in this exercice ; a = -4  b= -4  c = -1

then: the equation of a plane that is ;  -4x-4y-z +d =0

but the plane  passing through the point (−2,−5,5) :

-4(-2)-4(-5)-(5) +d =0

23+d =0

d =-23

the equation of a plane is : -4x-4y-z-23 =0

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slavikrds [6]
Since A (area of circle = C) is given = 148
Where we assume:
Y represents radius of the circle (r)
X represents diameter of the circle (D)
Pi (π) = 3.14
A = 2 * π * y
148 = 2 * 3.14 * y
148 = 6.28 * y
y = 148/6.28
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D = 2 * y
D = 2 * 23.56
So, D = 47.12

Assume A is unknown (not given as 148)
A = π * y^2
A = 3.14 * (23.56)^2
A = 3.14 * 47.12
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3 0
3 years ago
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3 0
3 years ago
Find the slope of the line that contains (3, –2) and (4, 3).<br> Can someone help fast please?
suter [353]

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7 0
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8 0
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3 0
3 years ago
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