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3 years ago

Graph the line whose y-intercept is 9 and whose x-intercept is -7

Mathematics

1 answer:

Tomtit [17]3 years ago

7 0

y-intercept = 9 → (0, 9)

x-intercept = -7 → (-7, 0)

Look at the picture.

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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:

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

5 0

3 years ago

Mia needs to bring ar least 25 pieces of fruit to the school picnic. She suggested both apples and oranges and was told to be su

SOVA2 [1]

I think x + y ≥ 25, x ≥ 10, y ≥ 0

7 0

3 years ago

Read 2 more answers

Help me please very confused

CaHeK987 [17]

Read the problem and answer choices. You want to get from ABCD to EFGH, so you need to figure out how to do that with reflection, translation, and dilation—in that order.

The reflection part is fairly easy. ABC is a bottom-to-top order, and EFG is a top-to-bottom order, so the reflection is one that changes top to bottom. It must be reflection across a horizontal line. The only horizontal line offered in the answer choices is the x-axis. Selection B is indicated right away.

The dimensions of EFGH are 3 times those of ABCD, so the dilation scale factor is 3. This means that prior to dilation, the point H (for example), now at (-12, -3) would have been at (-4, -1), a factor of 3 closer to the origin. H corresponds to D in the original figure, which would be located at (0, -2) after reflection across the x-axis.

So, the translation from (0, -2) to (-4, -1) is 4 units left (0 to -4) and 1 unit up (-2 to -1).

The appropriate choice and fill-in would be ...

... B. Reflection across the x-axis, translation 4 units left and 1 unit up, dilation with center (0, 0) and scale factor 3.

_____

You can check to see that these transformations also map the other points appropriately. They do.

8 0

3 years ago

Make a box-and-whisker plot for the data. What is the upper quartile value? 56 32 48 52 51 53 48 38 35 42 40 46 54 50

Alika [10]

From the figure the upper quartile is 51.75

7 0

4 years ago

Read 2 more answers

Solve for the distance between the points (4, 4) and (1, -3).

antoniya [11.8K]

I think the distance is 3/7. This is because 4 minus 1 is 3 (they are both in the same quadrant, so u subtract/are both positive too). -3 is three away from zero and 4 is 4 away from zero. If you add those together, you get those points (also, u add numbers in different quadrants together). Therefore, the distance is 3/7. Hope this helps!

7 0

3 years ago