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:
608 ft²
Step-by-step explanation:
<u>1) Find the area of the bases</u>
where b is the base length and h is the height
Plug in b and h

Multiply the answer by 2 (because there are 2 bases)
A=96
Therefore, the area of the two bases is 96 ft².
<u>2) Find the area of the two sides facing up</u>
where l is the length and w is the width
Plug in l and w

Multiply the answer by 2 (because there are 2 sides)
A=320
Therefore, the area of these two sides is 320 ft².
<u>3) Find the area of the bottom side</u>
where l is the length and w is the width
Plug in l and w

Therefore, the area of this side is 192 ft².
<u>4) Add all the areas together</u>
96 ft² + 320 ft² + 192 ft²
= 608 ft²
Therefore, the surface area of the triangular prism is 608 ft².
I hope this helps!
Width = 4ft
Length = 6ft
<u>Explanation:
</u>
Let length = l ,
So, width = 2l-8
Perimeter of rectangle = 2x(length + width)
According to the question,

Hence,
Width = 4ft
Length = 6ft
X+6=11
X-11=-6
5x=25
2x=10
2x+15=25