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Given the two parabolas y = 2 - x^2 and y = 5 + x^2, determine the equations of all common tangents.

Hallo guest!

\(determined\ graphically\)

\(f'_1(x)=2.4x+3.5\\ f'_2(x)=-2.4x+3.5\)

  !

The smallest 4-digit integer is 1,000. The largest is 9,992.

{9,992 - 1,000} / 8 + 1 =1,125 positive 4-digit integers divisible by 8

500 000/(1.05)¹⁰= 306956.63

Yep !

1- cos^2 + cos +1 = 0

cos^2 -cos -2 =0      let cos^2 = x^2    use quadratic formula  to find x(or cos x) = 2 or -1     x is cos x and cannot = 2   throw this value out

cos x = -1     at  180 degrees       

Can someone else verify this answer??

ok my answer still looks funky but is it like around $306956.63??

Q2  

x^2 - 6x - 2y + 7 = 0

Complete the square on 'x'

(x-3)^2 -9    - 2y+7 = 0      Move y and stuff to the right side of the equation

(x-3)^2 = 2y +2                  re-arrange to form

(x-3)^2 = 2 (y+1)

One of the roots of x^3 - 5x^2 + 5x - 1 = 0 is 2 - sqrt(3).  What are the other two roots?

Hallo Guest

\(I\ assume: x_1 = 1\)

 \((x^3 - 5x^2 + 5x - 1):(x-1)=\color{blue}x^2-4x+1\)

  \(\underline{x^3\ -\ x^2}\)

         \({-4x^2+5x}\)

         \(\underline{-4x^2+4x}\)

                          \(x-1\)

                          \(\underline{x-1}\)

                                  \(0\)

 \(x_1=1\)

 \(x^2-4x+1=0\\ x=2\pm\sqrt{4-1}\\ \)

\(x_2=2+\sqrt{3}\\ x_3=2-\sqrt{3}\)

  !

500 000 = PV (1.05)¹⁰       Solve for PV = amount to invest now

Second one

The center  is  found by the mid-point formula

(  [2 - 4]/ 2 , [7 + 4] / 2 )  =   ( -1, 11/2)

The radius  can be found by taking (1/2)  of the diameter...so

(1/2) sqrt   [ ( -4 - 2)^2  + ( 7 - 4)^2 ]  =  (1/2) sqrt [ (- 6)^2 + 3^2 ) =  (1/2) sqrt (45)  

r^2  =   45 / 4

So....the equation is

( x + 1)^2  + ( y - 11/2)^2   =  45 / 4

Here's a graph :  https://www.desmos.com/calculator/walfxegecn

Sorry...I mislead you just a bit....the 'n' should be a supescript....I fixed it above....do you see it should be  (1.05)ⁿ  

The form is    

(x - h)^2  +  ( y - k)^2  =  radius^2     where (h, k)  is the center.....so....

( x + 1)^2  + ( y -  7)^2  =  10^2

(x + 1)^2  +  ( y - 7)^2  = 100

3sin^2 x  = 8 cos x

3 ( 1 - cos^2 x)  = 8 cos x

3 - 3cos^2 x  = 8 cos x      rearrange as

3cos^2 x + 8cos x  -  3  = 0       let   cos x  =   m

3m^2  + 8m  - 3   = 0

(3m  - 1) ( m + 3)  = 0

The second factor has no real solutions for the  cosine

So

3m -  1  = 0

3m  = 1

m  =1/3

So

cos x = 1/3    take the  arccos

arccos  (1/3)  =  x  ≈ 70.53°  +  2pi * n        where n is an integer

And

x=  360 - 70.53   ≈  289.47° +  2 pi * n      where n is an integer

i don't think i got it from there :( here's what i thought:

now that i know all the variables(thanks to you!) i have:

500,000=(1.05)10p

500,000=10.5p

but then dividing both sides by 10.5 gives me a really funky number (47619.0476) so i don't think it's right.. what did i do wrong??

1st  iteration

f(12)  =  13/-11 =   -13/11

g (f(12))  =  -2  / [ -13/11 + 1 ]   =   -  2 / [ -2/11]   =  11

2nd iteration

f ( g ( f (12) ))   =   f (11)   =   12 / - 10  =   - 6/ 5

g ( f ( g ( f (12) )) )  =  g (-6/5)   = -2 / [ -6/5 + 1]   =  -2 / [ -1/5]  =  10

Continuing  like this

3rd iteration

f(10)  = -11/9

g (-11/9)  =  -2 / [ -11/9 + 1 ] =  -2 /[ -2/9]  =   9

So...it appears that each iteration  lessens the result   by  1

So

11,10, 9, 8 ,7, 6, 5, 4

So....after 8 iterations....we get    4

Another way is to use the QUadratic Formula....

you got pretty far

1/4 (x^2-6x+9) - 11

1/4 (x^2-6x-35)         use quadratic formula

a=1 b = -6 c = -35

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x=3±2sqrt(11)        Then continue as Cphil did.......don't forget the 1/4 in front....

f(x)   =  (1/4) ( x - 3)^2  - 11       

y  =  ( 1/4) ( x - 3)^2  -  11

y  =  (1/4)(x - 6x + 9)  -  11

y  =  (1/4)x^2 - (3/2)x  + 9/4  -  44/4

y  = (1/4)x^2 - (3/2)x  -  35 / 4

y =  (1/4)( x^2 - 6x  - 35)

a  =1/4

And we can solve this

x^2  - 6x  -  35  = 0       add   35 to both sides

x^2  - 6x   = 35          complete the square on x

x^2 - 6x + 9  =  35  + 9

(x - 3)^2   =   44   take both roots

x - 3  =   ±sqrt (44)

x - 3  =   ±2 sqrt (11)   add  3  to both sides

x =  3+ 2sqrt (11)    or    x =  3 - 2sqrt (11)

So....in factored form we have

y = (1/4)  [ x - ( 3 + 2sqrt (11) ) ]  [ x - (3 - 2sqrt (11) )   ]

See the graph here to verify  this  :  https://www.desmos.com/calculator/wx3tlpbvuk

FV = PV (1+.05)ⁿ

        FV = Future Value = 500 000    PV = Present Value = amount to invest now   .05 = interest rate   n = number of years=10

Can you take it from here?   (this assumes a one time investment now)

See  the following image   :

PQR  is isosceles  with  PQ  = PR.....so  QN  = (1/2) (QR)  = (1/2)(10)  = 5

By the Pythagorean Theorem

sqrt  ( QP^2  - QN^2)  =  PN

sqrt ( 13^2  - 5^2)  =  sqrt  ( 169 - 25)   = sqrt (144)  = 12 =  PN

Since  PQR is isosceles with PQ = PR then,  X  is a median......so

PX  =  (2/3) (12)   =    8

T(t)  =  T_A  +  ( T₀ -  T_A)e^(kt)

T_A  = room temp

T_{0  ⁼}  original temp of  roast

We need to find  k  , thusly.....we know that   that when t  = 10...we have.....

145  = 72  + ( 165 - 72)e^(k*10)       subtract 72 from both sides

73  =  (93) e^(k*10)      divide both sides by  93

(73/93)  = e^(k *10)       take the natural log   of each side

Ln  ( 73/93)  = Ln e ^(k *10)        and by a Ln property we can write

Ln (73/93)  = (k *10) Ln e          Ln e  = 1....so.....we can ignore this and we have

Ln (73/93)  = k * 10                 divide both sides by 10

Ln (73/93) / 10  = k   ≈   -0.0242

So....to find the time  it takes to  cool to  120°  we have

120  = 72 +  (93)e^(-0.0242 * t)       subtract  72 from both sides

48 =  93e^(-0.0242* t)     divide both sides by  93

(48/93)  =  e^(-0.0242 * t)     take the Ln of both sides

Ln  (48/93)  = Ln e^ (-0.0242 * t)         

Ln ( 48/93)  =  (-0.0242 * t) * Ln e         (ignore the  Ln e)

Ln (48/ 93)  = -0.0242 * t       divide both sides by  -0.0242

Ln (48 / 93) / [ -0.0242 ]  = t  ≈  27.33  minutes =   27 minutes

x^2          y^2

___  +    ____   =    1

50              9

The  eccentricity, e,   is given by  c  / a

c  = sqrt  [ a^2  - b^2 ]    =   sqrt  [ 50  - 9 ]  =  sqrt (41)

a  =  sqrt (50)

So

e  = sqrt (41) / sqrt (50)   ≈  .906  = .91

As e approaches 0,  we have  a more circular than eongated ellipse

As e approaches 1, we have  a more eongated than circular ellipse

So....this one is more elongated than circular

See the graph here   :  https://www.desmos.com/calculator/u6nvuljikf

what if you glued the egg to a piece of string/ribbon which was tied to a small helium-filled balloon? make the balloon the right size so that the egg doesnt float away but falls down very slowly and gently

is that allowed?

A= 8350(1+0.052/12)^10*12

A=8350(1 + 0.0043333)^120

A=8350(1.0043333)^120

A=8350(1.6801391)

A=$14,029.16

If x + y = 3, y + z = 5, and z + x = 4, then find x + y + z.

x +  y  =   3        (1)

y +  z  =   5        (2)

x + z  =    4        (3)

Subtract   (3)  from (2)   and we get that

y - x  =  1    ⇒    -x  + y  =   1     (4)

Add (1)  and (4)

2y  =  4

y  = 2

And   x  + y  = 3    implies that x  = 1

And  x + z =  4     implies that z  = 3

So

x + y +  z  =   1   + 2  +   3    =   6

its wrong :(((