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jok3333 [9.3K]
3 years ago
5

In this exercise, we show that the orthogonal distance d from the plane p with equation

Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0
Use the cross product to find the orthogonal vector, solve the parametric equation to see at which (t) the point + orthogonal vector intersects the plane, the distance is (t) * norm of vector
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What is the vertex of the function?
egoroff_w [7]
<h2>Explanation:</h2><h2></h2>

Here you haven't provided any function, so I'll assume this one:

f(x)=3x^2-18x+33

This is a quadratic function whose graph represents a parabola. Any parabola has its vertex at the point:

V(x,y) \\ \\ \\ x=-\frac{b}{2a} \\ \\ y=f(-\frac{b}{2a}) \\ \\ \\ where: \\ \\ \\ f(x)=ax^2+bx+c \\ \\ \\ So: \\ \\ a=3 \\ \\ b=-18 \\ \\ c=33

Therefore:

x=-\frac{(-18)}{2(3)}=3 \\ \\ f(3)=3(3)^2-18(3)+33=27-54+33=6

Therefore, the vertex is:

\boxed{V(3,6)}

6 0
3 years ago
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Norma-Jean [14]

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4 0
3 years ago
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Plz Help ASAP!
Rina8888 [55]
Part A:

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Part B: 
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Part C: 
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ALL of this is based on only looking at the graph, which is no better than drawing lines on the sand.. For example, the y intercept for g(x) could be 4.9 or 5.1, and I don't know the equation for f(x)... Based on the info I have,  gave my best answers.. 

Hope this helps.. 
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3 years ago
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Llana [10]

Answer:

Step-by-step explanation:

Hi,

I am quite good at maths and will be happy to help you, not just in Algebra, but any other topics in maths

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3 years ago
-5x - 5y + 4x - (2y + 3)
Dafna11 [192]

Step-by-step explanation:

-5x - 5y + 4x - (2y + 3)

= - 5x + 4x - 5y - (2y + 3)

= - x - 5y - 2y - 3

= - x - 7y - 3 ( Answer )

5 0
3 years ago
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