Rom

\(a_n = a_1 + (n-1)d\\ a_5 = a_1 + 4d = 9\\ a_{32} = a_1 + 31d = -84\\ \text{subtract }a_5 \text{ from }a_{32}\\ 27d=-93\\ d=-\dfrac{93}{27} = -\dfrac{31}{9}\\ a_1 = 9-4d = 9-4\left(-\dfrac{31}{9}\right) = \\ 9+\dfrac{124}{9} = \dfrac{205}{9}\\ a_{23} = a_1 + 22d = \dfrac{205}{9}+22\left(-\dfrac{31}{9}\right) = -53 \)

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You shouldn't have any trouble with the perimeter.  4 applications of the distance formula and sum.

For the area one way or the other you're going to need the height of the trapezoid.  That's not so easy.

As far as I can tell you're going to need to use the formula for the distance between two parallel lines.

\(\text{the top of the trapezoid has slope}\\ m=\dfrac{5-7}{6-3}=-\dfrac 2 3\\ (y_t-7)=-\dfrac 2 3 (x-3)\\ y_t = -\dfrac 2 3 x+9\)

\(\text{the bottom of the trapezoid has the same slope}\\ (y_b-5)=-\dfrac 2 3 (x+1)\\ y_b=-\dfrac 2 3 x +\dfrac{13}{3}\)

\(\text{The distance between two lines is}\\ d= \dfrac{|b_2-b_1|}{\sqrt{m^2+1}}\)

\(b_1=9,~b_2=\dfrac{13}{3},~m = -\dfrac 2 3\\ d = \dfrac{|9-\frac{13}{3}|}{\sqrt{\dfrac 4 9 + 1}} = \dfrac{\frac{14}{3}}{\sqrt{\dfrac{13}{9}}}=\\ \dfrac{14}{\sqrt{13}} = \dfrac{14\sqrt{13}}{13}\)

\(\text{the length of the top segment is}\\ \ell_t = \sqrt{(6-3)^2+(5-7)^2} = \sqrt{13}\\ \text{the length of the bottom segment is }\\ \ell_b = \sqrt{(-1-5)^2 + (5-1)^2}=\sqrt{52} = 2\sqrt{13}\)

\(\text{Now we can apply the area of a trapezoid formula}\\ A= \dfrac{\ell_t + \ell_b}{2} h = \\ \dfrac{\sqrt{13}+2\sqrt{13}}{2}\times\dfrac{14\sqrt{13}}{13} = \\ \dfrac{3\sqrt{13}}{2}\dfrac{14\sqrt{13}}{13} = 21\)

I kind of rushed through this and there may be an easier method.  I'm sure folks will chime in pointing out my errors.

\(f(g(x) = g(f(x))\\ f(g(x) = 3(ax+b)+2 = 3ax + (3b+2)\\ g(f(x)) = a(3x+2)+b = 3ax + (2a+b)\\ 3b+2 = 2a+b\\ 2b+2=2a\\ b=a-1\)

\(ab=20\\ a(a-1)=20\\ a^2-a-20=0\\ (a-5)(a+4)=0\\ a=5 \vee a=-4\\ 5+(-4)=1\)

we can reduce the the problem to the following

\((\text{length of the blue line})^2 =2^2 + (\text{length of the red line }-1)^2\)

\((\text{length of blue line})^2 = (t-2)^2 + (t-2)^2 = 2(t-2)^2\\ (\text{length of red line}-1)^2 =(\sqrt{ (t-3)^2+t^2}-1)^2 \\ \text{a modest amount of algebra gets us }\\ 6+2t = 2\sqrt{2 t^2-6 t+9}\\ 3+t=\sqrt{2 t^2-6 t+9}\\ 9+6t+t^2 = 2t^2 - 6t+9\\ 12t=t^2\\ t=12 \)

2 ways

1) click tan and hold it, the button will change to atan

2) click on the button labeled "2nd", the buttons will all change including the tan button

\(\text{the hole is created by }\dfrac{x-4}{x-4}\\ \text{the vertical asymptote is created by }\dfrac{1}{x-2}\\ \text{as we end up with a single power of }x \text{ in the denominator }\\ \text{the horizontal asymptote is created by }2x^2+x \\ \text{combining all this we have}\\ f(x) = \dfrac{(x-4)(2x^2+x)}{(x-4)(x-2)} = \dfrac{x(x-4)(2x+1)}{(x-4)(x-2)}\)

\(f(x) = \dfrac{2x^3-7x^2-4x}{x^2-6x+8}\)

\(\text{veranda }=320 cm\times 640cm\\ \text{we want to find the greatest common factor of }320 \text{ and }640\\ 320=5\times 2^6\\ 640=5 \times 2^7\\ \text{so the LCM of the two is }5\times 2^6 = 320\\ \text{we use square tiles }320cm \times 320cm = 3.2m \times 3.2m\\ \text{we'll need 2 tiles}\)

\(\text{I'm not seeing anything clever}\\ \begin{array}{lr} z_1=1+i &|z_1-i| = 1\not < \dfrac 1 2\\ z_2 = \dfrac{2}{2}+i = 1+i &|z_2-i|=1 \not < \dfrac 1 2 \\ z_3 = \dfrac{2}{3}+i &|z_3-i| = \dfrac 2 3 \not < \dfrac 1 2\\ z_4 = \dfrac{13}{36}+i &|z_4-i| = \dfrac{13}{36} < \dfrac 1 2 \end{array}\\ \text{the least value of }n \text{ such that }|z_n-i|<\dfrac 1 2 \text{ is }n=4\)

what you do with these is multiply top and bottom by the complex conjugate of the denominator

\(\dfrac{7+9i}{1+3i}=\dfrac{7+9i}{1+3i} \dfrac{1-3i}{1-3i} = \\ \dfrac{(7+27)+(9-21)i}{10} = \dfrac{34-12i}{10} = \\ \dfrac{17}{5} -\dfrac 6 5 i\)