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vaieri [72.5K]
3 years ago
13

Which Choice is equivalent to the quotient shown here when x is greater or equal to 0

Mathematics
2 answers:
m_a_m_a [10]3 years ago
6 0

\sqrt{ \frac{27x}{48} }  =  \sqrt{ \frac{ {3}^{3}x}{ {2}^{4} \times 3 } }  =   \sqrt{ \frac{ {3}^{2}x }{ {2}^{4} } }  =  \frac{3 \sqrt{x} }{4}
Licemer1 [7]3 years ago
5 0

Answer:

C

Step-by-step explanation:

using the rules of radicals

• \frac{\sqrt{a} }{\sqrt{b} } ⇔ \sqrt{\frac{a}{b} }

• \sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

given \sqrt{27x} ÷ \sqrt{48}

simplifying the radicals

\sqrt{27x} = \sqrt{9(3)x} = 3\sqrt{3x}

\sqrt{48} = \sqrt{16(3)} = 4\sqrt{3}, hence

\frac{3\sqrt{3x} }{4\sqrt{3} }

= \frac{3\sqrt{x} }{4} → C



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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:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

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erma4kov [3.2K]

Answer:

$3.75

Step-by-step explanation:

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C = 0.75

The cost per candle is $0.75

Multiply by 5 to get the cost of 5 candles

$3.75

Hope this was useful to you!

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Answer:
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Anastasy [175]

Answer:

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(see attached for reference on cosine rule)

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Put the dot on the 5 on the Y-axis


Step-by-step explanation:

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7 0
3 years ago
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