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melamori03 [73]
2 years ago
9

Find the coordinates of the midpoint of segment GH with endpoints: G(-2, -8) and H(-3, -12).

Mathematics
1 answer:
Musya8 [376]2 years ago
6 0

Answer:

midpoint = (-2.5 , - 10)

Step-by-step explanation:

midpoint =( ((x1 + x2)/2) , ((y1+y2)/2)) )

= ( ((-2 + -3)/2) ,(( -8 + -10)/2) )

= ( -2.5 , -10)

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Caroline is considering two video game rental plans. Plan A can be modeled with the equation C = 2n, and Plan B can be modeled w
meriva

Options

A. Caroline rents exactly 7 games each month.

B. Caroline rents exactly 6 games each month.

C. Caroline rents 6 or more games each month.

D. Caroline rents from 1 to 5 games each month.

Answer:

D. Caroline rents from 1 to 5 games each month.

Step-by-step explanation:

Given

Plan A:

C = 2n

Plan B:

C = n + 6

Required

Which options justifies A over B

The solution to this question is option (d).

In option d, n = 1,2,3,4,5

When any of the values of n is substituted in plan A and B, respectively; the cost of plan A is cheaper than plan B.

This is not so, for other options (A - C)

To show:

Substitute 1 for n in A and B

Plan A:

C = 2n  = 2 * 1 = 2

Plan B:

C = n + 6 = 1 + 6 = 7

Substitute 5 for n in A and B

Plan A:

C = 2n  = 2 * 5 = 10

Plan B:

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8 0
3 years ago
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}

5 0
3 years ago
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