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:
I think No
Step-by-step explanation:
The question is asking for the intersection of the line and the parabola. Where does the blue parabola and the red line meet? <em>at (-1, -3) and (3, 5)</em>
Answer: B) x = -1 , x = 3
I need a photo to tell you the answer
<h2>
Answer:</h2><h2>c = -24</h2><h2 /><h2>Hope this helps!!</h2>