CPhill

Show an equation of the tangent to the curve 𝑦 = 𝑥^5 − 3𝑥^2 +𝑥 + 5 at the point (−1, 0) is given by 𝑦 = 12(𝑥 +1) by differentiation

y' = 5x^4 - 6x + 1

The slope of the tangent line at (-1,0) is

y' (-1) =  5(-1)^4 - 6(-1) + 1      =    12

The equation of the tangent line to (-1,0) is

y = 12 ( x - 1) + 0

y = 12(x - 1)

Angle OQL = 90°

Angle OLQ bisects the interior angle MLQ =  120/ 2 =  60°

So...angle LOQ =  180 - 90 - 60 =   30°

CORRECTED......

The number of diagonals in a regular polygon of n sides = n(n - 3) / 2

So

n(n - 3) / 2 =  5n     multiply boh sides by 2

n(n -3) = 10n       simplify

n^2 - 3n = 10n     subtract 10n from both sides

n^2 - 13n = 0      factor

n ( n - 13) = 0

Set each factor to 0 and solve for n and we get

n= 0   [ reject]    or        n = 13

So....13 sides

Let the side length of the octagon = S.....this = the hypotenuse of an isosceles triangle

So...the length of one of the legs of the isosceles triangles =   S/√2

And the side of the square = 

S + 2S/√2 =  S + S√2

So

S + S√2  = 10

S(1 + √2)  = 10    divide bot sides by ( 1 + √2)

S =     10   / ( 1 + √2 )   ≈    4.14  (inches)

LOL!!!

Actually......your first answer was more "complete" if we allow y to be positive or negative....!!!

Remember that y > 0

(a + bi)^2  = 2( a - bi)

a^2 + 2abi - b^2 =  2a - 2bi

Equating the real and imaginary parts, we have that

(a^2 - b^2) = 2a              2abi = - 2bi

Using the second equation

2a = -2

a = -1

And using the first equation

(-1)^2 - b^2 = -2

-b^2 = -3

b^2 = 3

b = ±3

Notice that if y > 0

6 < y < 7...... will make this true

For example....if y = 6.1

The ceiling = 7

And the floor = 6

Here's what I find.....

Let a = m....the discriminant is

m^2 -  4*5 * (m)  = m^2 - 20m  = m(m - 20)

This will be a perfect square when m = 20   ⇒ a = 20

And

x^2  + 20x +5(20) =

x^2 + 20x + 100       has  the root (-10, 0) 

And if a = 0.....this has the root   (0, 0)  

So

a = 0   or  a = 20

Let A = (0,0)   B = (0, 4)   C = (4,4)   D = (4,0)

Let M = (4, 2)

Let the circle centered at A have the equation 

x^2 + y^2 = 16    (1)

Let the circle centered at M have the equation 

(x - 4)^2 + ( y - 2)^2 = 4

x^2 - 8x + 16 + y^2 - 4y + 4 = 4

x^2 - 8x + 16 + y^2 - 4y = 0   

x^2 + y^2 - 8x - 4y = 0   (2)

Subbing (1) into (2) we can find the y coordinate of P....this will be the distance from P to AD

16 - 8x - 4y  = 0

8x = 16 - 4y

2x = 8 - y

x = [ 8 - y] / 2   (3)

Sub (3) into (1)

[(8 - y)/2]^2 + y^2 = 16     simplify

[y^2  - 16y + 64] / 4 + y^2 = 16

y^2 - 16y + 64 + 4y^2 = 64

5y^2 - 16y = 0

y ( 5y - 16) = 0      set each factor to 0 and solve for y

y = 0  [ reject]             5y - 16  = 0

                                   5y = 16

                                     y = 16/5

This is the y coordinate of P.....so..it is 16/5 units from AD