A student is trying to solve the set of two equations given below:

1 answer:

7 0

If you would like to eliminate the z-term, you can do this using the following steps:

x + z = 6 /*(-2)
3x + 2z = 1
_____________
-2x -2z = -12
<span>3x + 2z = 1
</span><span>_____________
</span>-2x + 3x = -12 + 1
x = -11

The correct result would be <span>x + z = 6 /*(-2).</span>

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Find the focus and directrix of the parabola with the equation x² = -22y.

Bad White [126]

9514 1404 393

Answer:

focus: (0, -5.5)

directrix: y = 5.5

Step-by-step explanation:

The equation compares to the form ...

x² = 4py

where 4p = -22 ⇒ p = -22/4 = -5.5.

The value of p is the distance from the vertex to the focus. The negative value means the focus is below the vertex at (0, -5.5).

The directrix is above the vertex by the same amount, so is y = 5.5.

6 0

3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$