Bginner

Let $A$ be the number of terms in the expansion of $(x+a)^{11}$.  (This $A$ is also the number of terms in the expansion of $(x-a)^{11}$.)  Let $B$ be the number of terms in the expansion of $(x+a)^{11}$ that are the same as the terms in the expansion of $(x-a)^{11}$.  The answer is $A-B$.
 

Key idea:  : Let  y = Total paid for Plan A =  B ,   at x minutes use

we compare  two equations of plan A & B  and solve for x minutes use

Plan A :     y =   0.10 x 

Plan B:      y = (0.05)x + 20

0.10x  =  (0.05)x + 20

x =  400    

(Meaning:  Plan B will be cheaper after 400 min.  )

At x =  400  is the intercept between Plan A and B.  The slope of Plan B is increase slower than the Plan A

  Red line  y = (0.05)x + 20 . 

The $x^2$ term is $(-4x^2)(-7) + (5x)(tx) + (-5)(x^2)$ which is equal to $(28+5t-5)x^2$.  Since the product has no $x^2$, we have $28+5t-5=0$.  Solving for $t$ gives $t=-23/5$.

Linear function   y = mx +b   , 

y =  (3/4)  x -14   .....L₁

Let  m = slope of L₁  = (3/4)  

m₂ =  slope of  L₂

if the lines are perpendicular :     slope   ( m₁ * m₂ ) =  ( - 1 )   

   (3/4) m₂   = -(1) ,     m₂   =  (-1)* (4/3)  =   -(4/3)

Then L_2:   y =  -(4/3)x + b.......L₂

_{Find b:   Plug in (x = -5, y  = 7 ) into}    y = -(4/3)x +b.....L₂

Let $a$ denote the first term of the AP and let $d$ be the constant so that the $n$th term is $a+(n-1)d$.  The sum of the first five terms is $5\cdot(a + a+4d)/2 = 5\cdot(a+2d) = 20 +10d$.  The sum of the next five terms is $5\cdot((a+5d) + (a+9d))/2 = 5\cdot(a+7d) = 20 +35d$.  The former is one-fourth of the latter, so $20+10d=(20+35d)/4$, which solves to $d=-12$.  Therefore, the 15th term is $a+14d = 4-14\cdot 12=-164$.

It's 2.

Let the function $f$ be $f(x)=ax^2+bx+c$.   The zeroes of $f$ is the roots of the quadratic equation $x^2+\frac{b}{a}x+\frac{c}{a}=0$.  We are given that the zeroes of $f$ are at $x=2$ and $x=4$.  So the zeroes are the roots of the equation $(x-2)(x-4) = x^2-6x+8 = 0$.  We are also given that $f(3)=6$.  From these facts we get a system of three equations in three unknowns $a,b,c$:
\begin{eqnarray*}
    \frac{b}{a} &=& -6 \\
    \frac{c}{a} &=& 8 \\
    9a+3b+c &=& 6.
\end{eqnarray*}
This solves to $a=-6$, $b=36$, and $c=-48$.  Thus, $f(x)=-6x^2+36x-48$.

Let the function $f$ be $f(x)=ax^2+bx+c$.   The zeroes of $f$ is the roots of the quadratic equation $x^2+\frac{b}{a}x+\frac{c}{a}=0$.  We are given that the zeroes of $f$ are at $x=2$ and $x=4$.  So the zeroes are the roots of the equation $(x-2)(x-4) = x^2-6x+8 = 0$.  We are also given that $f(3)=6$.  From these facts we get a system of three equations in three unknowns $a,b,c$:
\begin{eqnarray*}
    \frac{b}{a} &=& -6 \\
    \frac{c}{a} &=& 8 \\
    9a+3b+c &=& 6.
\end{eqnarray*}
This solves to $a=-6$, $b=36$, and $c=-48$.  Thus, $f(x)=-6x^2+36x-48$.

Come to think of it, flipping should not be allowed.   Sorry.

The distance between a point $P=(x_0,y_0)$ and the line $L: ax+by+c = 0$ is $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.