Greetings!Solve for <em>x</em>.
Distribute the Parenthesis. <em>How?</em> Multiply all of the terms inside the parenthesis by the number outside the parenthesis.


Add
6y to both sides.


Divide both sides by
-2.
The Answer Is:


Hope this helps.
-Benjamin
The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

In this case, we have
<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )
<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2
and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then





Answer:They are equivalent, if that's what your trying to find out.
Step-by-step explanation:
Multiply by a across the formula