CPhill

x = y^4

x + y^2 = 1        sub the first into the second and we have that

y^4 + y^2  = 1        we can complete the square on y thusly

y^4 + y^2 + 1/4    = 1 + 1/4       factor and simplify

(y^2 + 1/2)^2 =  5/4           take both roots

y^2 + 1/2  =  ± sqrt (5) / 2

y^2 =  ±sqrt (5) / 2 - 1/2

y^2 =      [ -1 ± sqrt (5) ] / 2        take both roots again

y = ± sqrt [ [  -1 ± sqrt (5) ] / 2 ]

Let y  =   a ,  -a

So  ......x = y^4   =   a^4

So   we have the points      (a^4, a )   and ( a^4, - a)

So....the distance between these points  =

sqrt [ (a^4 - a^4) + ( a  -  -a)^2 ]  =  sqrt  [ (2a)^2]  =    2a

So...we have, 2a  =

2 √ [ ( -1  ± √5) / 2 ]      but we cannot take a root of a negative so the distance must be

2 √ [ ( -1  + √5) / 2 ]   =

√ [  4 ( -1  + √5) / 2 ]  =

√ [ - 2 + 2√5 ]

So  (u, v)  =  (-2, 2)

From the top, we have

2(4x+2)=4x−12(x−1)       simplify

8x + 4 =  4x - 12x + 12

8x + 4 = - 8x + 12         add 8x to both sides, subtract 4 from both sides

16x = 8                       divide both sides by 16

x = 8 /16  =  1 / 2      .....just as tertre found.....!!!

Can't draw the diagram.....but

tan 50° = height of statue / shadow length

tan 50° = height of statue / 30      multiply both sides by 30

30 * tan (50°) = height of statue  = 35.8 ft

May = 5th month.....so....x = 4

4.5 sin ( 4pi/6) + 11.8

4.5 sin (2pi/3) + 11.8

4.5 * sqrt (3) /2 + 11.8 ≈    15.7 C

I'm assuming that  e^2 - r is in the denominator

g / [ e^2 - r ]  = m - y         and we can write

[ e^2 - r ] / g  =    1 / [ m - y ]

e^2 - r =  g / [ m - y ]

e^2 =   g / [ m - y ]  + r

e =  ±√ [ g / [ m - y ] + r ]

e^(3x + 5) = 9

Ln e^(3x + 5)  = Ln 9

(3x + 5) Ln e = Ln 9

3x + 5  = Ln 9

3x =    -5 + Ln 9

x =  [ -5 + Ln 9 ] / 3

2^(x + 3) - 3

2^(x + 2) - 1

g(x) is f(x) shifted one unit to the right and  two units up

6x + 12

_____________

x^2  + 8x + 12

The denominator has three terms - True

2 is a common factor in num and den   - False

12 is a term in both num and den - True

x + 2  is a factor of the numerator - True

Look at my second answer to your other post, so1963  ...

If we have a right triangle   where one leg = 4x^2 cm and the other is 8x^2 cm

Then the area is

(1/2) 4x^2 * 8x^2  =

16x^4  cm^2