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2 years ago

If f (x) = StartRoot one-half x minus 10 EndRoot + 3, which inequality can be used to find the domain of f(x)?

Mathematics

2 answers:

Margaret [11]2 years ago

8 0

Answer:

C. One-half x minus 10 greater-than-or-equal-to 0

Step-by-step explanation:

just olya [345]2 years ago

7 0

The inequality that represents the domain of the function is $\frac 12x - 10 \ge 0$

<h3>How to determine the domain?</h3>

The function is given as:

$f(x) = \sqrt{\frac 12x - 10} + 3$

Set the radicand greater than or equal to 0

$\frac 12x - 10 \ge 0$

Multiply through by 2

x - 20 ≥ 0

Add 20 to both sides

x ≥ 20

Hence, the domain of the function is x ≥ 20

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3 years ago

state domain and range, intercepts, relative max and relative mins points and values, intervals of increase and decrease, interv

Aloiza [94]

The domain of the graph is -∝ < x < ∝ and the range of the graph is y > -2. Also, the symmetry is x = -3

<h3>The domain and range of the graph</h3>

From the graph, we can see that the x values extend indefinitely

So, the domain of the graph is -∝ < x < ∝

However, the minimum y value is -2 and it increases from there

So, the range of the graph is y > -2

<h3>The intercepts</h3>

This is where the graph crosses the axes

So, we have the intercepts to be

x-intercept = -1 and -5
y-intercept = 1

<h3>Relative max and relative mins points and values, </h3>

The graph has no relative maximum and the relative minimum is (-3, 2)

<h3>Intervals of increase and decrease</h3>

There are the intervals where the function increases and decreases

So, we have the intervals to be

Increasing interval: (-3, ∝)
Decreasing interval: (-∝, -3)

<h3>Intervals of positive and negative</h3>

There are the intervals where the function is positive and negative

So, we have the intervals to be

Positive interval: (-∝, -5) and (1, ∝)
Negative interval: (-5, -1)

<h3>The symmetry</h3>

This is the line that divides the graph equal part

In this case, the symmetry is x = -3

<h3>The end behavior</h3>

Using the intervals where the function is positive and negative, the end behavior is

As x ⇒ ∝, f(x) ⇒ +∝
As x ⇒ -∝, f(x) ⇒ -∝

Read more about functions at:

brainly.com/question/3951754

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2 years ago

A hyperbola is given by the equation x^2+2xy−y^2+x = 2. Use implicit differentiation

Mekhanik [1.2K]

$\bf x^2+2xy-y^2+x=2\qquad 
\begin{array}{llll}
(1&,&2)\\
x_1&&y_1
\end{array}\\\\
-------------------------------\\\\
2x+2\left( 1\cdot y+x\frac{dy}{dx} \right)-2y\frac{dy}{dx}+1=0
\\\\\\
2x+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}+1=0
\\\\\\
\cfrac{dy}{dx}(2x-2y)=-1-2x-2y\implies \cfrac{dy}{dx}=\cfrac{-1-2x-2y}{2x-2y}
\\\\\\
\left. \cfrac{dy}{dx}=\cfrac{2x+2y+1}{2y-2x} \right|_{1,2}\implies \cfrac{2+4+1}{4-2}\implies \cfrac{7}{2}\\\\
-------------------------------\\\\
$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=\cfrac{7}{2}(x-1)\implies y-2=\cfrac{7}{2}x-\cfrac{7}{2}
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
\boxed{y=\cfrac{7}{2}x-\cfrac{3}{2}}$

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4 years ago

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Olenka [21]

Answer:

Step-by-step explanation:

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6 0

3 years ago

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ioda

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3 years ago