This is a choice of two out of five without regard to order - that is, the number of combinations of two out of a set of five. This is

$\displaystyle C(5,2) = \frac{5!}{(5-2)!2!}= \frac{5!}{3!\; 2!} = \frac{120}{6 \cdot 2} = 10$

There are ten possible ways to choose, and $\displaystyle 10< 20$.

The square with the diagonal line alternates between the square on the left and the square on the right. Therefore, the next figure in the sequence will have its diagonal line in the rightmost square, eliminating Figures (b) and (d) and leaving Figures (a) and (c).

Also, the shaded square moves one position  to the right from figure to figure, so in the next figure, the shaded square must be the one at the extreme right. Figure (c) matches that description.

Use the point-slope formula with $\displaystyle m = -\frac{1}{5}, x_{1}= 1, y_{1} = 4$

$\displaystyle y -y _{1} = m (x -x _{1} )$

$\displaystyle y -4 = -\frac{1}{5} (x -1 )$

$\displaystyle y -\frac{20}{5} = -\frac{1}{5} x + \frac{1}{5}$

$\displaystyle y = -\frac{1}{5} x + \frac{21}{5}$

Rewrite each in the slope-intercept form, $\displaystyle y = mx + b$; $\displaystyle m$ will be the slope.

$\displaystyle 2x+y = 15$

$\displaystyle 2x+y -2x = 15-2x$

$\displaystyle y=15-2x$

$\displaystyle y = -2x+ 15$

The slope of this line is $\displaystyle m = -2$.

$\displaystyle 3x+y = 15$

$\displaystyle 3x+y -3x = 15-3x$

$\displaystyle y=15-3x$

$\displaystyle y = -3x+ 15$

The slope of this line is $\displaystyle m = -3$.

Since $\displaystyle -2 > -3$, (A) is greater.

Rewrite each in the slope-intercept form, $\displaystyle y = mx + b$; $\displaystyle m$ will be the slope.

$\displaystyle 4x - 2y = 10$

$\displaystyle 4x - 2y -4x = 10-4x$

$\displaystyle - 2y = -4x+ 10$

$\displaystyle - 2y \div (-2) =\left ( -4x+ 10 \right )\div (-2)$

$\displaystyle y = 2x -5$

The slope of the line of $\displaystyle 4x - 2y = 10$ is $\displaystyle m = 2$

$\displaystyle y - 2x = 7$

$\displaystyle y - 2x + 2x = 7+ 2x$

$\displaystyle y = 2x + 7$

The slope of the line of $\displaystyle y - 2x = 7$ is also $\displaystyle m = 2$

The slopes are equal.

The slope of a line through the points $\displaystyle (A+ 1, B+ 1 )$ and $\displaystyle (A-1, B- 1 )$ can be found by setting 

$\displaystyle x_{1} = A-1,y_{1} = B - 1, x_{2} = A+1, y_{2} = B+ 1$

in the slope formula:

$\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle = \frac{ (B+ 1)- (B- 1)}{ (A+ 1)- (A- 1)}$

$\displaystyle = \frac{ B-B+ 1+1}{ A-A + 1 + 1}$

$\displaystyle = \frac{ 2}{ 2}$

$\displaystyle =1$

The slope of a line through the points $\displaystyle (B- 1, A- 1 )$ and $\displaystyle (B+1, A+ 1 )$ can be found similarly:

$\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle = \frac{(A+ 1)- (A- 1) }{ (B+ 1)- (B- 1) }$

$\displaystyle = \frac{A-A + 1+1}{B-B + 1 + 1}$

$\displaystyle = \frac{ 2}{ 2}$

$\displaystyle =1$

The lines have the same slope.

The slope of a line through the points $\displaystyle (A, B)$ and $\displaystyle (-A, B )$, can be found by setting 

$\displaystyle x_{1} = -A,y_{1} =B, x_{2} = A, y_{2} = B$:

in the slope formula:

$\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle m = \frac{B-B}{A - (-A)} = \frac{0}{A+A} = \frac{0}{2A} = 0$

For this problem, simply plug in the values for the point (15,6) into the different equations (15 for the $\displaystyle x$-value and 6 for the $\displaystyle y$-value) to see which one fits.

$\displaystyle (6)=\frac{1}{2}(15)-7$          (NO)

$\displaystyle (6)=\frac{2}{3}(15)-4$          (YES!)

$\displaystyle (6)=\frac{-1}{2}(15)+7$       (NO)

$\displaystyle (6)=\frac{-2}{3}(15)+4$        (NO)  

At the intersection point of the two lines the $\displaystyle x$- and $\displaystyle y$- values for each equation will be the same. Thus, we can set the two equations as equal to each other:

$\displaystyle \frac{3}{4}x-11=\frac{-5}{6}x+8$

$\displaystyle \frac{3}{4}x=\frac{-5}{6}x+19$

$\displaystyle \frac{9}{12}x=\frac{-10}{12}x+19$

$\displaystyle \frac{19}{12}x=19$

$\displaystyle (\frac{12}{19})(\frac{19}{12}x)=19 (\frac{12}{19})$

$\displaystyle \large x=12$

$\displaystyle y=\frac{3}{4}(12)-11= -2$

point of intersection $\displaystyle =\left ( 12,-2 \right )$

In the formula $\displaystyle y=mx+b$, the y-intercept is represented by $\displaystyle b$ (because if you set $\displaystyle x$ to zero, you are left with $\displaystyle y=b$ ).

Thus, to find the $\displaystyle x$-intercept, set the $\displaystyle y$ value to zero and solve for $\displaystyle x$.

$\displaystyle 0=\frac{7}{9}x-3$

$\displaystyle \frac{-7}{9}x=-3$

$\displaystyle (\frac{-9}{7})\frac{-7}{9}x=-3 (\frac{-9}{7})$

$\displaystyle x=\frac{27}{7}=3\frac{6}{7}$