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\(R(\theta) = \begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}\\ \text{Now just plug in $\theta=\dfrac{\pi}{6}$}\)

it's not

(-1)^{(any even number)} = 1

we would need to divide the scale value (1.6 *10^16)  we choose by all the values , thats the table i was talking about

The table is in your question, JL !

Do you have the table with all the values ? Just so I can make more sense of it ? Thank you again ! You have been such a big help 

thank you!

Looks like Chris is signed off presently.....

Looking for b ?

a^2 = b^2 + c^2

a = 58  (million miles)

c = 12  (million miles)

58^2 = b^2 + 12^2

b = 57   million miles

Here is the orignal post 

https://web2.0calc.com/questions/calculus-help_15

Go ahead and post it.....the distance from the horizontal endpoint to where  =  46   ???

You're close, CU......this actually is a parabola that opens to the right

We can find the y intercepts by setting  x  = 0  and solving

y^2 - 4y  - 1   = 0

y^2 - 4y  =   1        complete the square on y

y^2 - 4y + 4  =   1 + 4

(y - 2)^2  =  5      take both roots

y - 2 = ±√5

y  =  2 ±√5

So.....there are two y intercepts  as shown by CU's  graph....!!!!

oh no! the distance from the horizontal endpoint is 46 and for the other side its 70 miles add them together to get the full horizontal diamater 116. sorry for the confusion im not fully understanding the question to be honest. if it helps i could post the whole question. 

I'm assuming that the  distance from the center to the focus  =  46   =  c

116/2  =  a  = 58

In an ellipse

b^2 + c^2  = a^2        ....so

b^2  = a^2  - c^2

b^2  = 58^2  - 46^2

b^2  = 3136  - 2116

b = √[ 3136 - 2116 ]  = √1020 = 2√255

OOPS!!!....thanks, EP!!!

Use   T = time in SECONDS

         G = gravitational constant    6.67408 x 10^-11

         M = mass of Sun   1.989 x 10³⁰

And the equation     T =  sqrt  (4 pi² r³ /(GM)  )       Solve for r   which will be in METERS

Then divide by AU (in meters )   1.4959787 x 10¹¹                  Let me know what ya find for an answer.....it IS in your list of possibles !

nvm its (0,0)

what is my origin/center?

f(x) = y = 100 x

The shortest one is 2 m      2m  x 100cm/m = 200 cm

similarly the other two are 300 cm and 400 cm

domain would be the lenghts        2 <= x <= 4

Range would be the cm lengths   200<=y<=400

Let's put the circle in standard form  

x^2 + 4x +y^2 -6y = 3    now 'complete the square'on x and y

(x+2)*2 + (y-3)^2 = 3  +4+9

(x+2)^2 + (y-3)^2 = 4^2           so the center   is  (a,b) = (-2,3)   and the radius is 4       a+b+r = 5

Here is a graph of the rectangle (the 4th corner is  5,-2)   and the equation of the circle....

I think you can see the area in question is 1/4 the area of the circle:

https://www.desmos.com/calculator/eno5sm5r56

= [3(-2)]2   -  4(-3)2

= 36 - 4(9) = 0

yes thank you , 

for the second part would i have to transform general into standard?