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Elis [28]
3 years ago
11

I need help RIGHT NOW PLEASE HELP NOW!!!!ldentify the vertex of the graph. Tell whether it is a minimum or maximum. A (1-2); min

imum B (1 -2); maximum C (2,1); minimum D (-2, 1); maximum

Mathematics
1 answer:
larisa [96]3 years ago
3 0

Answer:

Answer D:  (-2, 1):  maximum

Step-by-step explanation:

This graph is that of a parabola that opens down.  The graph clearly has a maximum.  

That max is at x = -2 and y = 1:  (-2, 1) (Answer D).

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4 years ago
QUICK PLEASE HELP ME!!!!!
11111nata11111 [884]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
Divide 7/24 by 35/48
Degger [83]

Answer:

0.4 or 4/10 as a fraction

Step-by-step explanation:

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3 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

x=-2(2)=-4

y=2

z=2

With\lambda =- 2

x=-2(-2)=4

y=-2

z=-2

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

7 0
3 years ago
Find the 13th term of the sequence 4, 8, 16, 32, .
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Answer:

The thirteen number will be 16384.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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