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Setler [38]
2 years ago
11

Help with this please.

Mathematics
1 answer:
Ivanshal [37]2 years ago
8 0

Answer:

  • True
  • True

Step-by-step explanation:

This question asks you to compare the coordinates of the vertex of each function.

__

The vertex of the function is its minimum, the point where the graph stops decreasing and starts increasing. It is the lowest point on the graph.

<h3>f(x)</h3>

The vertex is (-4, -1). The minimum is -1, located at x = -4.

<h3>g(x)</h3>

The vertex is (1, -25). The minimum is -25, located at x = 1. We know this is the minimum because there are no g(x) values that are lower (more negative).

<h3>comparison</h3>

The minimum of f(x), -1, is greater than the minimum of g(x), -25. TRUE

The x-value of f(x) at its minimum, -4, is less than the x-value of g(x) at its minimum, 1. TRUE

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Answer:

a) no real solutions

Step-by-step explanation:

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barxatty [35]

z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}

We're given that f_x(6,1)=3 and f_y(6,1)=-1, and want to find \frac{\partial z}{\partial v}(1,2).

By the chain rule, we have

\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}

and

\dfrac{\partial x}{\partial v}=2v

\dfrac{\partial y}{\partial v}=-1

Then

\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)

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4 0
3 years ago
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Answer:

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∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

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c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))

∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))

d) ∀x¬∃y (P(x, y) → Q(x, y))

∀x∀y ¬(P(x, y) → Q(x, y))

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3 0
2 years ago
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nignag [31]

Answer:

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I hope this helps you:))

P.S. Might I recommend a good online calculator for these types of problems. It's called Symbolab. I use it every day and it's a lifesaver. It shows the step-by-step and has a plethora of different actions like simplifying, inverse, and line.

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