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2 years ago

I do not know this answer

Mathematics

1 answer:

Bogdan [553]2 years ago

4 0

Answer:

try (-4,6)

Step-by-step explanation:

if you use tracing paper, put a mark where the point is

put the tip of the pencil on the point which it will rotate around

spin the paper by the degrees

see where the traced point ends up

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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

Helllllpppppppppppppppppp

Bogdan [553]

The sum of the inner angles of a polygon with n sides is $180(n-2)$

Since your polygon has 6 sides, the sum of its inner angles has to be

$180(6-2) = 180 \cdot 4 = 720$

On the other hand, the sum of its inner angles is

$75+132+113+108+160+x = 588+x$

And it has to be 720, so we have

$588+x=720 \iff x = 720-588 = 132$

7 0

3 years ago

Apply the distributive property to produce an equivalent expression. 12 + 20

notsponge [240]

Answer:

10 + 20 + 2

Easy, make it easy to do mentally than add the remaining

8 0

3 years ago

Read 2 more answers

Point Kis on line segment JL. Given JL = 4x + 2, KL=5x– 6, and JK = 3x, determine the numerical length of JK.

Salsk061 [2.6K]

Answer:6

Step-by-step explanation:

6 0

3 years ago

-8(6-r)=24 what is the solution for r

Andreas93 [3]

Answer:hi

Step-by-step explanation:Let me first warn you that I CANNOT get the denominators of this problem to go underneath the numerators. It looks good on my answer page but when I click "preview your answer" the denominators are ALL over the place. Grrrrr.

I think the weirdly placed denominators make this explanation hard to understand but you give this a go and see if you understand things. If it is too hard to see, maybe this website will let you instant message me or email me. If so, then I will send you the solution via email. I'm very sorry the denominators are SO scattered.

See what you think of the explanations below but it might help if you write down my answer on paper, so that you can move the number to rest more closely under their proper numerators. :-)

Ok then......

Let's start with your first problem. A quick way to do this work is to cross multiply. Here's what I mean:

__3___ = __6___

r - 1 r + 2

Ok, so you can cross multiply:

3(r + 2) = 6(r - 1) (cross multiply)

3r + 6 = 6r - 6 (use the distributive property)

Now you will have to put the "r" variables on the left side of the equal sign and numbers without variables on the right side of the equal sign. To get 6r from the right side to the left, you can subtract 6r from the right side. If you subtract it from the right, you must subtract it from the left.....

To get 6 from the left side to the right, you can subtract it from the left. If you subtract it from the left, you must subtract it on the right.....

So now you have:

3r - 6r = -6 - 6 This becomes.........

- 3r = -12

Now to solve for "r" you will have to divide by -3. Why? Well, -3r is really saying: -3 times r. The -3 and the r are actually being multiplied together so to isolate the "r" you have to sort of "undo" the multiplication. What is the opposite of multiplication? It's DIVISION. Soooo, you must divide by -3 on both sides. SO you have:

- 3r = -12

r = 4 (-12 divided by -3 is a positive 4)

Now you know "r" equals 4. If we "plug" 4 back into the original equation, we can see the answer is correct. Check this out:

__3___ = __6___ (original equation)

r - 1 r + 2

__3___ = __6___ (substitute "4" in for "r"

4 - 1 4 + 2

__3___ = __6___ (do the math in the denominator area)

3 6

1 = 1 (3/3 equals 1 and 6/6 also equals 1, so our

solution that r = 4 is correct!) :-)

Now to your 2nd problem..........

4/2x-5 = 24/2x

___4___ = __24___

2x � 5 2x

Let's cross multiply again:

4(2x) = 24(2x - 5)

8x = 48x - 120

8x - 48x = - 120

-40x = - 120

(divide both sides by -40)

x = 3

Now let's "plug" 3 in where we had the variable "x" in the original equation to see if we are correct.

___4___ = __24___ (original equation)

2x � 5 2x

___4___ = __24___ (substituted 3 for "x")

2(3) � 5 2(3)

___4___ = __24___

6 - 5 6

___4___ = __24___

1 6

4 = 4

SO once again we have found the correct answer. Yay for us! :-)

I hope this helps. :-)

8 0

3 years ago

Read 2 more answers