Bginner

Since $P(x)$ is cubic and $x^2+x+1$ is quadratic, the quotient must be linear.   Let the quotient be $ax+b$ for some constants $a$ and $b$.
Then
$$P(x) = (x^2+x+1)(ax+b) + 2x+3.$$
We are given that $P(0)= -3$, so
$$(0^2+0+1)(a(0)+b) + 2(0)+3 = -3$$
that is,
$$b+3=-3,$$
giving $b=-6$.
We are also given that $P(1) = 4$, so
$$(1^2+1+1)(a(1)-6) + 2(1)+3 = 4$$
that is,
$$3(a-6)+5=4$$
giving $a=\frac{17}{3}$.
Therefore, the quotient is $\frac{17}{3}x-6$.

Let constant = C

then 2048 * C = 128

C   =  (128/2048)  =   (1/16)

x = 128 * (1/16) = ?

y = (x* (1/16) = ?

Let f(x) = 1/(x - 8).  Find the largest x that is not in the domain of g(x) = f(f(x)).

Key idea:    Rational function : The denominator cannot be zero, any x value that makes denominator equal zero is not in the domain.

\(f(f(x)) = \frac{1}{(\frac {1} {x-8}) -8}\)

\(g(x) = \frac{1}{\frac{65-8x}{x-8}}\)

\(g(x) = \frac{x-8}{65-8x}\)

If (65 -8x = 0) that x value is not the domain of g(x)

65-8x = 0

x = (65/8) = 8.125    the largest x that is not in the domain of g(x)

Key idea:

1) If the given sequence does have common difference, or common ratio, then it is neither an arithmatic nor geometric.

2) How to find out :

We say "d" is common difference of arithmatic :

If d value is   (a2 - a1) ≠ (a3 - a2 )  then it is not arithmatic sequences.

We say "r" is common ratio of geomatric sequences:

If r value is  (a₁ / a ₂) ≠ (a₂ / a₃ ) , then it is not geomatric sequences. 

...  Here, we get common ratio of geomatric sequence =  (1/4) /(1/6) = (1/6) /(1/9) = (3/2)  , Then it is geometric sequences.

3) Is  (3/2) greater  than 1?

4) If geometric common ratio  is greater than 1 , it means the sum of the series  keep increasing , then we say it is diverges.

\(f (x) = {1\over |(x-1) (x+4)+|(x-1)(x+10)|}\)

if x = 1, then

\(f(x) = {1 \over0}\)

f(x)  is undefined,  when x = 1.

Note: This is a correction of the previous message.   I don't know how to delete a message!

See the wiki page https://en.wikipedia.org/wiki/Factorial_number_system for information on the Factorial Number System.

First, we write the base-10 number 60 in the Factorial number system.  We calculate
 

    $60 \div 1 = 60$, remainder 0

    $60 \div 2 = 30$, remainder 0

    $30 \div 3 = 10$, remainder 0

    $10 \div 4 = 2$, remainder 2

    $2 \div 5 = 0$, remainder 2

This gives $60_{10} = 2:2:0:0:0_!$
Thus, the 60th permutation of 1, 2, 3, 4, 5 in lexicographic order is $34125$

Substituting the given values into the length formula
$(x_1-x_2)^2 + (y_1-y_2)^2 = \ell^2$ we get
$(2a-4)^2 + ((a-4)-(-7))^2 = (\sqrt{18})^2$.
This can be simplified to $a^2-2a+\frac{7}{5}=0$.
We have $\alpha, \beta$ are the roots of the quadratic equation
$(a-\alpha)(a-\beta)=0$ if and only if $a^2-(\alpha+\beta)a+\alpha\beta=0$.
Therefore, the product of the possible values of $a$ is
$\alpha\beta = \frac{7}{5}$.

See the wiki page
https://en.wikipedia.org/wiki/Factorial\_number\_system
for information on the Factorial Number System.
First, we write the base-10 number 60 in the Factorial
number system.  We calculate
 

    $60 \div 1 = 60$, remainder 0

    $60 \div 2 = 30$, remainder 0

    $30 \div 3 = 10$, remainder 0

    $10 \div 4 = 2$, remainder 2

    $2 \div 5 = 0$, remainder 2

This gives $60_{10} = 2:2:0:0:0_!$
Thus, the 60th permutation of 1, 2, 3, 4, 5 in lexicographic order
is $23145$.

Key idea:

1) Slope of the perpendicular line is oposite recipocal slope to the other

line equation  y  = mx + b  ,

perpendicular line y = (- 1/m)x  + b.

or

[Formula:  oposite recipocal    m₁ * m₂ = -1].

2) The given line equation  y = (6/5)x -30 .....(1).

slope of the given line = 6/5

Then oposite recipocal to (6/5)  =  negative of *(5/6) = - (5/6)

3) Get the  perpendicuar line equation by using the given point ( x= -6,  y= -7)

4)  y = mx + b    ,  -7 = (-5/6)x +b .....(2)

5) Find the  b value :  where x = 0   in (2)   .  Then  b =  -7.

The perpendicular line y = (oposite recipocal slope) *x + b

My answer is 79.238 m²

(square area - area of the 30 degree sector)

(96 -16.762 = 79.238)

1) Area of a sector of circle =  (area of a circle )* (Degree angle of the sector / 360)

..[30/360 = 1/12 is like the sector is one part of the whole  12 parts of the whole area of (pi r ² )]...

(pi * 8²  ) * (1/12)  =  16.7619m²

2) The dry area of the garden :  Garden area  - 16.7619  =   79.238 m²