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a)   (x - 2)^2  + (y - 1)^2   = 13        (1)

Second part.... expand (1)   and we have

x^2 - 4x + 4    + y^2 - 2y + 1  = 13     simplify

x^2 + y^2 - 4x - 2y  +  5   = 13       subtract 13 from both sides

x^2 + y^2 -4x - 2y - 8  =   0 

b)  A  = (4, k)

Sub 4 into (1)  for x

(4 - 2)^2  + ( y - 1)^2  = 13

2^2  + (y - 1)^2  = 13

4 + (  y - 1)^2  = 13

(y - 1)^2  = 9       take both roots

y - 1  = ±3

y =  ±3 + 1

So  y  =  3 + 1  = 4     or  y = -3 + 1  =  -2

So..... k  =  4     or k = -2

c) The tangent will meet the circle's radius at a  right angle...so...we have the right triangle ABC

AC will be one leg of the triangle  = sqrt (13)

BC will form the hypotenuse  = 7

And AB will form the other leg

So....using the Pythagorean Theorem

AB  =  sqrt  [  7^2  -  (sqrt (13) )^2  ]   = sqrt [ 49 - 13 ] = sqrt (36)  =  6

The function can't be a quadratic.

a)

EP 's first answer in red  ⇒   y  =  4 / x^2

It crosses the x axis at x  = - 1

The...it  crosses the x axis again at x = 1

And f(-2) =  f(2)

So.....this appears to be a quadratic  [ a parabola that turns upward ]  =  degree 2

Here is another way!

1 =1/1,000,000 x (1 +i)^1,000      divide both sides by 1/1,000,000

1,000,000 = (1 +i)^1,000               take 1/1,000th root of both sides

1,000,000^(1/1,000) = 1 + i

1.013911385........     = 1 + i           subtract 1 from both sides and multiply by 100

i =1.3911385%.

hmm i see but what about a)?

Y ~~ 1/x^2         y = k/x^2

when y = 4 and x = 1        4 = 1/(1^2)  * k     so k = 4

y=4/x^2

25 = 4/x^2       x^2 = 4/25     x = 2/5

FV = PV (1+i)^n

1 = 1x10^-6 (1+i)^1000

1/1x10^-6 = (1+i)^1000

10^6 = (1+i)^1000                  take log of both sides

6 = 1000 log(1+i)

6/1000 = log (1+i)       raise 10 to both sides

 10^(6/1000) = 1+i

i=1.3911%

Welcome aboard   !!!!!

Thanks, Dimitristhym.......!!!!

Nice CPhill! 

In the equation of the circle.....sub   2x + 11   for y......and we have

( x + 5)^2  +  (  2x + 11 - 6)^2   =  25      simplify

x^2 + 10x  +  25   + ( 2x + 5)^2   =  25

x^2 + 10x + 25 +  4x^2 + 20x + 25  = 25

5x^2 + 30x + 50  = 25        subtract 25 from both sides

5x^2 + 30 x + 25   = 0         factor out 5

5 ( x^2 + 6x + 5)   =  0       divide both sides by 5

x^2 + 6x + 5   = 0

Factor  x^2 + 6x + 5  =  0     as

(x + 5) ( x + 1)  = 0

Set each factor to 0  and solve for x and we have

x = - 5      and  x  =  - 1

These are the x coordinates of the intersections...and the y coordinates are

y = 2(-5) + 11  = 1        and    y = 2(-1) + 11  =  9

So...the intersection points are

(-5, 1)   and ( -1, 9)

Correct.....the cosine  and sine   are negative in Q3   !!!!!

We know that   x  = 8  and y = -7

r  =  sqrt  [  (8^2   + (-7)^2 ]   =   sqrt [ 64 + 49]  =  sqrt [ 113]

So.....cos (theta)   =  x / r   =   8 / sqrt (113)

Multiply both numerator and denominator by  sqrt [ 113]   and we have

8 sqrt (113)                             8sqrt (113)

_________________ =         __________

sqrt(113) * sqrt (113)                 113

Thanks, tertre   !!!

2.

P dilated by a factor of  (1/3) about the origin  =   (12, 6) (1/3)  =  (4, 2)

Q dilated by a factor of 1/3 about the origin  =  (-9, 15)(1/3)  =   ( -3, 5)

No prob, RP  !!!!

Let's work with this, first

  2               1

___    +     ___

x - 1            x

Cross-multiply  and keep the same sign as we have between the fractions

2x +   1 * ( x  - 1)   =   2x  + x - 1   =    3x - 1

This wiill be the numerator of the fraction at the top

Take the product of the denominators = x ( x - 1)

This will be the denominator of the fraction at the top

So.....we have

3x - 1                8

______   ÷       ____                invert the second fraction  and multiply

x(x - 1)               x

3x - 1                x                   x ( 3x - 1)

_______    *  _____  =      ____________

x ( x - 1)           8                 8x ( x - 1)

Cancel the x's on top/bottom

(3x - 1)

______

8 (x - 1)

Thank you for your help!

Note that we have a common denominator at the top....so this is just

[ 1 - x^2  ] / 4

Notice that  x  can be wriiten as    x  / 1

So  we have

[ 1 - x^2 ] / 4   ÷  [ x / 1 ]

Invert the second fraction and multiply

( 1 - x^2)               1                   (1 - x^2)

_______    *      ____   =          _________

      4                    x                        4x

tan x  = 15 / 8

To find the angle x, take the arctan  of 15/8

arctan (15/8)  ≈  61.93°

Thank you! ^-^ 

Lets work with this, first

3                  1

____    -     ___

a^2             3

Cross-multiply ans keep the same sign as between the fractions

3*3 - 1*a^2   =  9 - a^2

Take the product of the denominators 

3a^2

So   we have

( 9 - a^2) 

________

    3a^2

So...puitting this together, we have

[ a^2 / 9 ]  ÷  [ ( 9 - a^2) / ( 3a^2) ]

Invert the second fraction and multiply

a^2               3a^2                      3a^4

___   *         ______    =           _______   

9                  9 - a^2                  9(9 - a^2)

Cancel  the 3 in the numerator and the 9 in the denominator  and we are left with

a^4                        a^4

_______     =      ________

3(9 - a^2)           27 - 3a^2