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2 years ago

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.

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

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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Last one plz help me!

Mashutka [201]

Answer:

a) no real solutions

Step-by-step explanation:

the discriminant (b² - 4ac) is negative which indicates no real solutions

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3 years ago

Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa

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

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

(because the point $(x,y)=(6,1)$ corresponds to $(u,v)=(1,2)$ )

$\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}$

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3 years ago

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT(x, y,

Gnesinka [82]

Answer:

a) ∀x∃y ¬∀zT(x, y, z)

∀x∃y ∃z ¬T(x, y, z)

b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]

∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

∀x∀y [¬P(x, y) ^ ¬Q(x, y)]

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))

∀x∀y (¬P(x, y) ^ Q(x, y))

3 0

2 years ago

Read 2 more answers

What is the standard form and the factored form

Feliz [49]

Answer:

See below

Step-by-step explanation:

<u>The factored form, use variables/numbers outside the square:</u>

(x + 6)(x - 4)

<u>The standard form, use variables/numbers inside the square:</u>

x² + 6x - 4x - 24 = x² + 2x - 24

3 0

2 years ago

Simplify the polynomial (3a^2+4a+25) - (2a^2+4a-5)

nignag [31]

Answer:

The polynomial simplified would be a^2 + 30

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.

6 0

3 years ago