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3 years ago

1000000×345671234567

Mathematics

2 answers:

Tcecarenko [31]3 years ago

7 0

Answer:

The answer is 345671234567000000. Gosh, im so smart

Anything that has a 1 followed by any number of zeros (for example: 3x10=30 or 4x100=400, see you're just adding a zero.)

Anarel [89]3 years ago

5 0

Answer:

Step-by-step explanation:

i think that answer is 345671234567000000

have a nice day

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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

2 years ago

A car is traveling at a rate of 72 kilometers per hour. What is the car's rate in kilometers per minute? How many kilometers wil

nataly862011 [7]

Answer:

The car’s rate in kilometer per minute is 1.2 kilometer/minute and in 20 min it can travel 24 kilometer

Solution:

It is given that it can travel 72 kilometer per hour.

We know 1 hour = 60 min

So, the car travels in

60 min = 72 hour

$1 \min =\frac{72}{60} \text { hours or } \frac{6}{5} \text { hours or } 1_{5}^{1} \text { hours. }$

Thus the cars rate in kilometer per minute is $\frac{6}{5}$ is 1.2 kilometer per minute.

Now we know in 1 minute the car can travel 1.2 kilometer.

Therefore in 20 minutes it will travel 1.2 $\times$ 20 kilometers= 24 kilometers

5 0

2 years ago

2.State whether ABC and LAED are congruent. Justify your answer.

sladkih [1.3K]

Answer

C. Yes, by SSS only

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How to solve this part

san4es73 [151]

Try this solution (the way shown is not the shortest one):

The task on the left part:

1. note, the point P and R have coordinates (-4;-4) and (-3;-2).

2. using the coordinates of points P and R it is possible to make up the equation of the required line:

$\frac{x-x_P}{x_R-x_P} =\frac{y-y_P}{y_R-y_P}; \ => \ \frac{x+4}{-3+4}=\frac{y+4}{-2+4}; \ => \ \frac{x+4}{1}=\frac{y+4}{2} \ or \ y=2x+4$

The task of the right part:

1. note, the common view of the equation of the line is y=kx+b, where b - an intersection point of the Y-axis and the given line; k=QR/PQ.

2. according the picture b= -5, k=4/1=5, so y= -4x-5

4 0

3 years ago

Could someone help me find the x and y

Maslowich

This triangle is a 45 - 45 - 90 triangle, making it an isosceles triangle. So, x is 13. In every isosceles right triangle, the hypotenuse is the length of each leg times the square root of 2. So, y is 13 times the square root of 2

5 0

3 years ago