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2 years ago

14

Find the domain and range of f(x ) = (x^2+2x+1)/(x^2-8x+12)

1 answer:

andrey2020 [161]2 years ago

8 0

Nter any function to compute the domain<span> and </span>range<span>. Specify a restriction on the independent variable. Compute </span>domain<span> and </span>range<span> for functions of several variables. ... </span>domain<span> and </span>range<span> of (</span>x^2+1)/(x^4-1) ...domain<span> of </span>f(x,y) = log(1-(x^2+y^2<span>)).</span>find<span> the </span>range<span> of </span>f(x<span>) = </span>x2+2x+1/x2-8x+12<span>. ... Write </span>domain<span> and </span>range<span> of greatest integer function. hopes this helps u

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These tables represent a quadratic function with a vertex at (0, -1). What is

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Answer:

B

Step-by-step explanation:

We can notice that based on the given pattern, the average rate of change from $x=n$ to $x=n+1$ , where $n \in \mathbb{Z}$ , is $-2n-1$ .

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3 months ago

Find the obtuse angle between the following pair of straight lines.

ioda

Given:

$y=-2$

$y=\sqrt{3}x-1$

To find:

The obtuse angle between the given pair of straight lines.

Solution:

The slope intercept form of a line is

$y=mx+b$ ...(i)

where, m is slope and b is y-intercept.

The given equations are

$y=0x-2$

$y=\sqrt{3}x-1$

On comparing these equations with (i), we get

$m_1=0$

$m_2=\sqrt{3}$

Angle between two lines whose slopes are $m_1\text{ and }m_2$ is

$\tan \theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$

Putting $m_1=0$ and $m_2=\sqrt{3}$ , we get

$\tan \theta=\left|\dfrac{\sqrt{3}-0}{1+(0)(\sqrt{3})}\right|$

$\tan \theta=\left|\dfrac{\sqrt{3}}{1+0}\right|$

$\tan \theta=\pm \sqrt{3}$

Now,

$\tan \theta= \sqrt{3}$ and $\tan \theta=-\sqrt{3}$

$\tan \theta= \tan 60^\circ$ and $\tan \theta=\tan (180^\circ-60^\circ)$

$\theta= 60^\circ$ and $\theta=120^\circ$

$120>90$ , so it is an obtuse angle and $60$ , so it is an acute angle.

Therefore, the obtuse angle between the given pair of straight lines is 120°.

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2 years ago