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Regular octagon \(ABCDEFGH\) has a side length of 2 cm.

Find the area of square \(ACEG\).

\(\text{ Let the side length of the square $=s$ } \\ \text{ Let the area of the square $=s^2$ } \\ \text{ Let $\angle ABC =135^\circ \qquad 135^\circ = 180^\circ-\dfrac{360^\circ}{8}$ } \)

cosine formula:

\(\begin{array}{|rcll|} \hline s^2 &=& 2^2+2^2-2\cdot 2\cdot 2\cdot \cos(135^\circ) \\ s^2 &=& 8-8\cdot \cos(135^\circ) \\ && \boxed{ \cos(135^\circ) \\ =\cos(180^\circ-45^\circ)\\ = -\cos(45^\circ)\\=-\dfrac{\sqrt{2}}{2} } \\ s^2 &=& 8-8\cdot \left( -\dfrac{\sqrt{2}}{2}\right) \\ \mathbf{s^2} &=& \mathbf{8+4\cdot\sqrt{2}} \\ \mathbf{s^2} &=& \mathbf{13.6568542495\ cm^2} \\ \hline \end{array}\)

The area of square \(ACEG\) is \(\mathbf{\approx 13.7\ cm^2}\)

list them

I maintain that 15 and 36 are the correct answers.

Do the division of this fraction to find    the result (3x-1) =m      since x cannot equal 1    m cannot equal  (3(1)-1)    or  2

From the given equation x cannot equal  -2   (that would result in a zero denominator)  so D= -2

If yo do the division of this fraction you wind up with   x^2+6x+9       A = 1   B = 6   C = 9    and D = -2

   I think you an take it from here.....

Hectictar thank you so much! Sorry if i wasted your time. Your solution is clear, smart, and correct!

Mmm.

Thanks Alan, seems I oversimplified.  I'll have to take a proper  look. 

1) 

\(\text{The remainder can be any degree $0-4$}\)

2)

\(\text{by the polynomial remainder theorem}\\ p(-1)=5\\ p(-5)=-7\\~\\ p(x)\pmod{(x+1)(x+5)} = ax+b\\ p(x) = q(x)(x+1)(x+5) + (ax+b)\\ p(-1) = 5 = 0+(-a+b)\\ p(-5) = -7 = 0 + (-5a+b)\\ 4a= 12\\ a=3\\ b=8\\~\\ p(x) \pmod{(x+1)(x+5)} = 3x+8\)

Here is a SIMILAR Q from yesterday! :

https://web2.0calc.com/questions/help_50407

The  curve is shifted UPWARD by  6     (&    The AMPLITUDE is 12)

The midline will be   y=6 

What does EP say? He says : AC =14 x cos 33 =14 x 0.838670... =11.74 cm.

There are values \(A\) and \(B\) such that \(\dfrac{Bx-11}{x^2-7x+10}=\dfrac{A}{x-2}+\dfrac{3}{x-5}\).
Find \(A+B\).

\(x^2-7x+10 = (x-2)(x-5)\)

\(\begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \quad & | \quad \times (x-2)(x-5) \\\\ & Bx-11 &=& A(x-5) + 3(x-2) \\\\ \hline x=5: & 5B-11 &=& A(5-5) + 3(5-2) \\ & 5B-11 &=& 0 + 3*3 \\ & 5B-11 &=& 9 \\ & 5B &=& 9+11 \\ & 5B &=& 20 \\ & B &=& \dfrac{20}{5} \\ & \mathbf{ B} &=& \mathbf{4} \\ \hline x=2: & 2B-11 &=& A(2-5) + 3(2-2) \\ & 2B-11 &=& -3A + 0 \\ & 3A &=& 11-2B \quad &|\quad B=4 \\ & 3A &=& 11-2\cdot4 \\ & 3A &=& 11-8 \\ & 3A &=& 3 \quad &|\quad :3 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline \end{array}\)

\(\mathbf{A+B} = 1+4 \mathbf{= 5}\)

so AC = 32 ?

The ladder is the hypotenuse of a right triangle      One leg is 20 feet   

sin = opp / hyp  

sin 32 = opp /hyp

hyp = opp/sin32 =  37.74 ft

14 cos 33 = AC        (cos = adjacent / hypotenuse )

If \(\sqrt{5 + x} + \sqrt{20 - x} = 7\), what is the value of \((5 + x)(20 - x)\)?

\(\begin{array}{|rcll|} \hline \sqrt{5 + x} + \sqrt{20 - x} &=& 7 \quad &|\quad \text{square both sides} \\ \left(\sqrt{5 + x} + \sqrt{20 - x}\right)^2 &=& 7^2 \\ \left(\sqrt{5 + x}\right)^2 +2\sqrt{5 + x} \sqrt{20 - x} + \left(\sqrt{20 - x}\right)^2 &=& 49 \\ 25 +2\sqrt{(5 + x)(20 - x)} &=& 49 \quad &|\quad -25 \\ 2\sqrt{(5 + x)(20 - x)} &=& 49-25 \\ 2\sqrt{(5 + x)(20 - x)} &=& 24 \quad &|\quad : 2 \\ \sqrt{(5 + x)(20 - x)} &=& 12 \quad &|\quad \text{square both sides} \\ \left( \sqrt{(5 + x)(20 - x)}\right)^2 &=& 12^2 \\ \mathbf{(5 + x)(20 - x)} &=& \mathbf{144} \\ \hline \end{array}\)

By looking at it    x = 4

5+4   *   20-4

9      *    16      =  144

Approach this by first drawing circles centred on point (-5,0) of radii 1, 2, 3 ... 10.

Then draw circles centred on point (5,0) with the same radii.

At the points where these circles intersect the distance from A is a positive integer less than or equal to 10, as is the distance from B.

It is a simple matter to count these intersection points. There are, indeed, 119 of them.

Thank you!

F = the temperature

decrease by 33 1/3 %

F * (1-.33333)       now increase by 50 %

(F * (1-.33333) (1.50) = F         No net change at all....right back to the original temp, F

x+1 =-2(y^2+10y+25)

x = -2y^2-20y-51               x will equal 0 at the y intercepts

0 = -2y^2-20y-51              now use the Quadratic Formula to find the y's

\(y = {-b \pm \sqrt{b^2-4ac} \over 2a}\)        a = -2   b = -20   c = -51       Let me know what you find !

                                                                           (Edit:  'cause when I graph it, I find NO y-axis intercepts!)

Find the remainder when the polynomial \(\large{x^{1000}}\) is divided by the polynomial \((x^2+1)(x+1)\). 

\(\begin{array}{|lrcll|} \hline &\mathbf{ x^{1000} }&=& \mathbf{(x^2+1)(x+1)q(x) +\underbrace{ax^2+bx+c}_{=r(x)}} \\ \hline x=-1:& (-1)^{1000} &=& \left((-1)^2+1\right)(-1+1)q(x) + a(-1)^2+b(-1)+c \\ & 1 &=& 0 + a-b+c \\ &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ \hline x=i:& (i)^{1000} &=& (i^2+1)(i+1)q(x) + ai^2+bi+c \quad | \quad \boxed{i^2=-1\\i^{1000}=\left(i^2\right)^{500}=\left(-1\right)^{500}=1} \\ & 1 &=& 0 -a+bi+c \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ \hline x=-i:& (-i)^{1000} &=& \left((-i)^2+1\right)(-i+1)q(x) + a(-i)^2+b(-i)+c \quad | \quad \boxed{(-i)^2=-1\\ (-i)^{1000}=1} \\ & 1 &=& 0 -a-bi+c \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline (2)+(3): & -a+bi+c+-a-bi+c &=& 1+1 \\ &-2a+2c&=& 2 \quad &| \quad :2 \\ &- a+ c&=& 1 \\ & \mathbf{c} &=& \mathbf{1+a} \qquad (4) \\ \hline (1) : & a-b+c &=& 1 \quad &| \quad \boxed{c = 1+a} \\ & a-b+1+a &=& 1 \\ & 2a-b &=& 0 \\ & \mathbf{b} &=& \mathbf{2a} \qquad (5) \\ \hline (2) : & -a+bi+c &=& 1 \quad &| \quad \boxed{b = 2a,\ c = 1+a}\\ & -a+2ai+1+a &=& 1 \\ & 2ai &=& 0 \quad &| \quad :2i \\ &\mathbf{a} &=& \mathbf{0} \\ & b &=& 2a \\ & b &=& 2*0 \\ &\mathbf{b} &=& \mathbf{0} \\ & c&=& 1+a \\ & c &=& 1+0 \\ &\mathbf{c} &=& \mathbf{1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r(x) &=& ax^2+bx+c \\ r(x) &=& 0*x^2+0*x +1 \\ \mathbf{r(x)} &=& \mathbf{1} \\ \hline \end{array}\)

The remainder is 1

In the diagram, \(D\) and \(E\) are the midpoints of \(\overline{AB}\) and \(\overline{BC}\) respectively.

Determine the sum of the \(x\) and \(y\) coordinates of \(F\), the point of intersection of \(\overline{AE}\) and \(\overline{CD}\).

\(\text{Let $\vec{A} =\dbinom06 $} \\ \text{Let $\vec{D} =\dfrac{\vec{A}}{2} = \dbinom03 $} \\ \text{Let $\vec{C} =\dbinom80 $} \\ \text{Let $\vec{E} =\dfrac{\vec{C}}{2} = \dbinom40 $} \\ \text{Let $\vec{F} = \dbinom x y $} \)

\(\begin{array}{|lrcll|} \hline (1) & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \quad \lambda \text{ is a scalar } \\ (2) & \vec{F} &=& \vec{D}+\mu(\vec{C}-\vec{D}) \quad \mu \text{ is a scalar } \\ \hline &\vec{F} = \vec{E}+\lambda(\vec{A}-\vec{E}) &=& \vec{D}+\mu(\vec{C}-\vec{D}) \\ & \mathbf{ \vec{E}+\lambda(\vec{A}-\vec{E}) } &=& \mathbf{\vec{D}+\mu(\vec{C}-\vec{D})} \\ & \lambda(\vec{A}-\vec{E})-\mu(\vec{C}-\vec{D}) &=& \vec{D}-\vec{E} \\\\ &&& \boxed{ \vec{A}-\vec{E} =\dbinom06-\dbinom40=\dbinom{-4}{6} \\ \vec{C}-\vec{D} =\dbinom80-\dbinom03=\dbinom{8}{-3} \\ \vec{D}-\vec{E} =\dbinom03-\dbinom40=\dbinom{-4}{3} }\\\\ & \lambda\dbinom{-4}{6}-\mu\dbinom{8}{-3} &=& \dbinom{-4}{3} \quad | \quad \times \dbinom{3}{8} \\\\ & \lambda\dbinom{-4}{6}\dbinom{3}{8}-\mu\underbrace{\dbinom{8}{-3}\dbinom{3}{8}}_{=0} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda\dbinom{-4}{6}\dbinom{3}{8} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda(-12+48) &=& -12+24 \\ & \lambda(36) &=& 12 \quad | \quad : 12 \\ & 3\lambda &=& 1 \\ &\mathbf{ \lambda } &=& \mathbf{ \dfrac{1}{3} } \\ \hline & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \\ & \vec{F} &=& \dbinom40+\dfrac{1}{3}\dbinom{-4}{6} \\ & \vec{F} &=& \begin{pmatrix} 4-\dfrac{4}{3} \\ \dfrac{6}{3} \end{pmatrix} \\ & \mathbf{\vec{F}} &=& \mathbf{ \begin{pmatrix} \dfrac{8}{3} \\ 2 \end{pmatrix} } \\\\ & x+y &=& \dfrac{8}{3} + 2 \\ &\mathbf{ x+y } &=& \mathbf{\dfrac{14}{3}} \\ \hline \end{array}\)

1      sum of squares is 1         

1 1      sum of squares is 2            

1 2 1       sum of squares is 6           

1 3 3 1       sum of squares is 20         

1 4 6 4 1       sum of squares is 70        Edited, Thanks Heureka :))