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3 years ago

Solve the system of linear equations below by graphing. x-y=17 and 3x+y=3

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Answer:

x=5,y=-12

Step-by-step explanation:

You add the two equations to get: 4x=20

Then you get x=5.

You plug this in to x-y=17 to get y=-12

The answer is x=5 and y=-12

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Distance between A(2,7) and B(6,4)

r-ruslan [8.4K]

Answer:

Approx: 2.65units

Step-by-step explanation:

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3 years ago

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Collect data from all your family members and write down their heights (in cm).

bija089 [108]

Answer:

mode : the value that appears most frequently in a data set

mean : te everage of all number

median : the middle number in a sorted, ascending or descending

range : the difference between the largest and smallest numbers.

ANSWER ;

mode : none

mean : 145.4 ( you need to add all the number and divide by how many your family member)

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3 years ago

Radium-226 is a metal with a half-life of 1600 years. If you start with 1 kilogram of radium-226, how much will remain after 20,

erica [24]

Ending Amount = Beginning Amount / 2^n
where 'n' = # of half lives
n = 20,000 / 1,600 = 12.5 half-lives

Ending Amount = 1 kg / 2^12.5
Ending Amount = 1 kg / 5,792.6
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4 years ago

For a field trip for students Road in cars and the rest filled nine buses how many students were in each bus it for how to 72 st

jasenka [17]

72 divided by 9 equals 8
72/9=8

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4 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