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Xelga [282]
3 years ago
10

The LCM of 13 and 9

Mathematics
1 answer:
babymother [125]3 years ago
4 0

Answer:

The Least Common Multiple of 13 and 9 is 13×9=117 because 13 is prime so they share no factors to create multiples of.

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-20

Step-by-step explanation:

-20

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Neal's room has sides 10 1/4 ft and 12 5/6 ft. What is the perimeter?
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Find the answer using Perimeter formula.

Step-by-step explanation:

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my yearly income is 18,000. My medical insurance premium is 230 a month. Is my medical premium 9.66% less then yearly income?
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Find an equation of the tangent plane to the given parametric surface at the specified point.
Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

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