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sergij07 [2.7K]
3 years ago
14

Find the missing measurement.

Mathematics
2 answers:
laiz [17]3 years ago
6 0

Answer: 9.6 km

Rounding of to nearest tens- 10KM

Serjik [45]3 years ago
6 0
10 when you round it but the first answer would be 9.6
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You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

6 0
2 years ago
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
2 years ago
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