CPhill

x^2 - 3y^2  = 13

 x - 2y  =  1  

Rearrange the second equation as

x  = 2y + 1

Sub this into  the first equation for x

(2y + 1)^2 - 3y^2  =  13      simplify

4y^2 + 4y + 1  - 3y^2  = 13

y^2 + 4y + 1  = 13     subtract 1 from both sides

y^2 + 4y - 12  =  0      factor

(y + 6) (y - 2)  = 0

Setting each factor to 0 and solving for y we have

y + 6   =  0              y  - 2  = 0

y  = -6                     y = 2

And when y  = -6 ....  x  = 2(-6) + 1  =  -11

And when y  = 2.....  x  = 2(2) + 1   =  5

So....the solutions are

(x,y)  =  ( 5, 2)  and  ( -11, -6)

(1/2)  to base 3

(1/2) * 3   =  3/2     ⇒  1  + 1/2

(1/2) * 3  =   3/2     ⇒  1  + 1/2

(1/2) * 3  =  3/2    ⇒   1  +  1/2

........

Read  the numbers  in red from bottom to top =  .111...₃

(1/7) base  8  

(1/7) * 8  =  8/7   =  1  +  1/7

(1/7) * 8  = 8/7    =  1  + 1/7

(1/7) * 8  = 8/7   =    1 + 1/7

........

Read the red numbers from bottom to top  =  .111...₈

So  it appears  that   1  / (b - 1)  in  base b  is  .111...._.b

We can orient the dish so that the vertex  is at the origin and the parabola opens upward...

If the dish  is  60cm wide  and 30cm deep, the points  (-30, 30)  and  (30, 30)  are on the graph

So......the form we have when the vertex  is  at the origin is this :

4py =  x^2    where p  is the distance between the origin and the focal point

And since   (30,30)  is on the graph, we have

4p (30)  = 30^2

120p = 900        divide both sides by 120

p = 900/120

p = 15/2

So.....the cooking grill should be located  at  (0, 0 + p)  =   (0, 15/2)

This fits the definition of the parabola because any light waves striking the interior of the dish will be reflected towards the focal point.

The function is

4 (15/2)y  =   x^2 ⇒  30y  =  x^2  ⇒   y  =  (1/30)x^2

Here  is the graph   with the focal point  :

https://www.desmos.com/calculator/r3x9j7epfp

1. First eight terms  ..... (−500)+(−100)+(−20)+(−4)+(−0.8)+...

The common ratio is  (1/5)....the sum, S, is given by :

S  =  First Term  [  1 -  common ratio^8] / [ 1 - common ratio ]  =

S  =  -500  [  1  - (1/5)^8] / [ 1 - 1/5]   =  -[ 624  + 624/625 ]  ≈  -624.99

2.  What is the sum of the first 7 terms of the series  −3+6−12+24−... ?

The common ratio  is  -2

The sum, S, is given by :

S  =  -3  [ 1  - (-2)^7 ] / [  1 - (-2) ]   = -129

1. 

\(f(x) = (\sqrt 5)^x \)

What is the range of  f(x)  on \([0, \infty)\)

This function is continuous on the requested interval

At  x  =  0, f(x)  = 1

Since √5  > 1, this function  is constantly increasing  on [0, inf)

So....the range   is   [1, infinity )  on  [0, infinity )

This is a 30-60-90 right triangle....the side across from the 30° angle  =12

The side across from the 60° angle   (what we are interested in ) is √3  times  this

So

12√3    =   20.78   ft

Note  cos B  = adjacent side  / hypotenuse  = a / c

So

cos B   = 6 / 11

Take the arccos  (cosine  inverse) to find  B

arccos   (6/11)  = 56.944268849146°

Degrees  =  56

Minutes  = 60 * .944268849146   =  56.65613094876  =   57 minutes

So

B  = 56° 57'

P (D l M)  = P(D and M) / P(M)        rearrange as

P ( D and M )  =  P (D l M)  * P (M)

P (D and M)  =   0.7  *  0.6  =      0.42

If A  =  60°.....B  =  30°

This is a  30-60-90  right triangle.....the side across from the 30°  is   1/√3  as long as the side opposite the  60° angle

So

b  = 15 * 1/√3  ≈   8.66

a  = 10

b = 4

Just a simple  Pythagorean Triangle problem  ;

c  = √ (10^2  + 4^2 )   =  √ (100 + 16)  = √ 116  ≈  10.77