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fgiga [73]
3 years ago
6

The point D(-3, 1) is reflected over the x-axis. What are the coordinates of the resulting point, D′?

Mathematics
1 answer:
PtichkaEL [24]3 years ago
6 0

Answer:

The rule for reflecting over the X axis is to negate the value of the y-coordinate of each point, but leave the x-value the same.

Step-by-step explanation:

For example, when point P with coordinates (5,4) is reflecting across the X axis and mapped onto point P', the coordinates of P' are (5,-4).

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2 years ago
A cylinder has a height of 14 centimeters and a radius of 17 centimeters. What is its volume? Use ​ ≈ 3.14 and round your answer
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<h2><u>Answer</u></h2>

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2 years ago
Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa
Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) :x^2+y^2+4z^2=36

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

\nabla f(x,y,z)=\\\nabla f(x,y,z)=

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)

So, 2x=-4\lambda

\Rightarrow x=-2\lambda

2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda

Substitute the value of x , y and z in the ellipsoid equation

(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2

With \lambda = 2

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y=2

z=2

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Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

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Amira is solving this problem.
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I think its B im pretty sure feel free to mark me brainlyest it would mean alot. Hope this helped!
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2 years ago
Read 2 more answers
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