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
![\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cmathrm%20ds%20%3D%20%5Cint_a%5Eb%20%5Csqrt%7B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%2B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
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
![\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7B%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5E2%2B%28-2%29%5E2%7D%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D-2%2Be%5E%7B-2t%7D%2B4%7D%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D%2B2%2Be%5E%7B-2t%7D%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Csqrt%7B%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5C%2C%5Cmathrm%20dt)
![=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}](https://tex.z-dn.net/?f=%3D%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5Cbigg%7C_0%5E2%20%3D%20%5Cleft%28e%5E2-e%5E%7B-2%7D%5Cright%29-%5Cleft%28e%5E0-e%5E%7B-0%7D%5Cright%29%20%3D%20%5Cboxed%7Be%5E2-%5Cfrac1%7Be%5E2%7D%7D)
Answer:
cosine = adjacent/hypotenuse
cos A = 20/29 (choice: yellow)
Step-by-step explanation:
<em>Hey</em><em>!</em><em>!</em>
<em>1</em><em> </em><em>and </em><em>2</em><em> </em><em>are</em><em> </em><em>vertical</em><em> </em><em>angles</em><em>.</em>
<em>Vertically</em><em> </em><em>opp</em><em>osite</em><em> </em><em>angles</em><em> </em><em>are</em><em> </em><em>always</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>each</em><em> </em><em>other</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em>
To write an equation you need a slop and y intercept. To find the slop use the formula y2-y1 /x2-x1. and for the y intercept use the formula y=mx+b
Answer:
35 and 7/12 feet is the combined height of the two records.
Step-by-step explanation:
Height of the indoor pool vault record for men = 20 and 1/6 ft =![\frac{121}{6} ft](https://tex.z-dn.net/?f=%5Cfrac%7B121%7D%7B6%7D%20ft)
Height of indoor pool vault record for women = 15and 5/12 ft =![\frac{185}{12} ft](https://tex.z-dn.net/?f=%5Cfrac%7B185%7D%7B12%7D%20ft)
The combined height of the two records :
![\frac{121}{6} ft+\frac{185}{12} ft](https://tex.z-dn.net/?f=%5Cfrac%7B121%7D%7B6%7D%20ft%2B%5Cfrac%7B185%7D%7B12%7D%20ft)
![\frac{121\times 2}{6\times 2} ft+\frac{185}{12} ft](https://tex.z-dn.net/?f=%5Cfrac%7B121%5Ctimes%202%7D%7B6%5Ctimes%202%7D%20ft%2B%5Cfrac%7B185%7D%7B12%7D%20ft)
![\frac{242}{12} ft+\frac{185}{12} ft](https://tex.z-dn.net/?f=%5Cfrac%7B242%7D%7B12%7D%20ft%2B%5Cfrac%7B185%7D%7B12%7D%20ft)
![\frac{242+185}{12} ft=\frac{427}{12}=35\frac{7}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B242%2B185%7D%7B12%7D%20ft%3D%5Cfrac%7B427%7D%7B12%7D%3D35%5Cfrac%7B7%7D%7B12%7D)
35 and 7/12 feet is the combined height of the two records.