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3 years ago

Find a vector function that represents the curve of intersection ofthe two surfaces. The cylinder

2 answers:

4 0

You're already given $x(t)=6\cos t$ and $y(t)=6\sin t$ . To find the $z$ component, plug these into the second equation:

$z=xy=36\sin t\cos t=18\sin(2t)$

so we have

$\vec r(t)=6\cos t\,\vec\imath+6\sin t\,\vec\jmath+18\sin(2t)\,\vec k$

Readme [11.4K]3 years ago

3 0

Answer:

here'

Step-by-step explanation:

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Number two please <br> ASAP

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The solutions for $-5 +2x^{2} = -6x$ is option 3. $x= \frac{-6 \pm \sqrt{36-4 (2)(-5)}}{4} .$

Step-by-step explanation:

Step 1:

First, we must bring the equation to the form of $ax^{2} +bx +c =0.$

So $-5 +2x^{2} = -6x$ becomes $2x^{2} +6x-5=0.$

The value of a is the coefficient of the $x^{2}$ term, the value of b is the coefficient of x term and c is the coefficient of the constant term.

Comparing the above equation to $ax^{2} +bx +c =0,$ we get $a = 2, b = 6,$ and $c=-5.$

We have the formula $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.$

Step 2:

By substituting the known values, we get

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} =\frac{-6 \pm \sqrt{6^{2}-4 (2)(-5)}}{2 (2)}.$

$\frac{-6 \pm \sqrt{6^{2}-4 (2)(-5)}}{2 (2)} = \frac{-6 \pm \sqrt{36-4 (2)(-5)}}{4} .$

This is the third option.

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4 years ago

5 more than 20 what is 5 more than 35 <br> What is five more than 50 what is five more than 85

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