I need to calculate the speed of light can someone give me an equation for doing so?

Mr. Garcia went on a long drive. When he started, there were 9 gallons of gas in his tank. After he had driven 26 miles, there w

Lyrx [107]

Answer:

y=-1 x=26 (26,-1)

Step-by-step explanation:

We know that Mr. Garcia drives 26 miles per gallon, this means that the slope will be (26,-1)

6 0

2 years ago

A construction elevator can carry at most 2,320 pounds. If the elevator operator weighs 195 pounds and each palette of roofing m

zvonat [6]

The maximum number of palettes that the elevator can carry in one trip is 5.

425 times 5= 2,125+ 195= 2,320

Hope this helps!! :-)

6 0

2 years ago

Which functions are correctly graphed? Select all the correct answers

garri49 [273]

the second one, and the third one

4 0

2 years ago

B is the midpoint of AC. If Ab= x +5 and BC = 2x-11, find the measure of Ab.

LuckyWell [14K]

__________________________

Measurement of "AC" :

(x + 5) + (2x − 11) ;
________________________

Find the measurement of "AB" [which is: "(x+5)" ]:
______________________________________________
First, simplify to find the measurement of "AC" :
________________________________________
(x + 5) + (2x − 11) ;

= (x + 5) + 1(2x − 11) ;

= x + 5 + 2x − 11 ;

→ Combine the "like terms" ;

x + 2x = 3x ;

5 − 11 = - 6 ;
______________
to get: 3x − 6 ;
_______________
So, (x + 5) + (2x − 11) = 3x − 6 ;
_______________________________
Solve for: "(x + 5)"
_______________________________

We have:
_______________________________

(x + 5) + (2x − 11) = 3x − 6 ;

Subtract: "(2x − 11)" ; from EACH SIDE of the equation ;
to isolate "(x + 5)" on one side of the equation;
and to solve for "(x + 5)" ;
________________________________________________________
→ (x + 5) + (2x − 11) − (2x − 11) = (3x − 6) − (2x − 11) ;

→ (x + 5) = (3x − 6) − (2x − 11) ;
_________________________________________________
Note: Simplify: "(3x − 6) − (2x − 11)" ;

→ (3x − 6) − (2x − 11) ;

= (3x − 6) − 1(2x − 11) ;

= 3x − 6 − 2x + 11 ;
__________________________
→ Combine the "like terms" :
_____________________________
+3x − 2x = 1x = x ;

-6 + 11 = 5 ;
_____________________________
To get: x + 5 ;

So we have:
______________________________
x + 5 = x + 5 ;
______________________________

So, x = all real numbers.

x = ℝ

4 0

2 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

2 years ago