The \(\displaystyle y\)-intercept of the graph of an equation is the point at which it intersects the \(\displaystyle x\)-axis. Its \(\displaystyle x\)-coordinate is 0, so set \(\displaystyle x = 0\) and solve for \(\displaystyle y\):

\(\displaystyle y = \frac{3}{5x-8}\)

\(\displaystyle y = \frac{3}{5 (0)-8}\)

\(\displaystyle y = \frac{3}{0-8}\)

\(\displaystyle y =- \frac{3}{8}\)

\(\displaystyle \left ( 0, -\frac{3}{8}\right )\) is the \(\displaystyle y\)-intercept.

The graph of \(\displaystyle y = \frac{6}{8x-2}\) does not have an \(\displaystyle x\)-intercept. If it did, then it would be the point on the graph with \(\displaystyle y\)-coordinate 0. If we were to make this substitution, the equation would be

\(\displaystyle 0= \frac{6}{8x-2}\)

and 

\(\displaystyle 0 \cdot (8x-2)= \frac{6}{8x-2} \cdot (8x-2)\)

\(\displaystyle 0=6\)

This is identically false, so the graph has no \(\displaystyle x\)-intercept. Therefore, the line cannot exist as described.

Substitute \(\displaystyle \frac{4}{x}\) for \(\displaystyle y\) in the second equation:

\(\displaystyle x+3\cdot \frac{4}{x} = 4\)

\(\displaystyle x+ \frac{12}{x} = 4\)

\(\displaystyle x+ \frac{12}{x}- 4 = 4 - 4\)

\(\displaystyle x- 4 + \frac{12}{x}= 0\)

\(\displaystyle \left (x- 4 + \frac{12}{x} \right )\cdot x = 0 \cdot x\)

\(\displaystyle x^{2}- 4 x + 12 = 0\)

The discriminant of this quadratic expression is \(\displaystyle b^{2} - 4ac\), where \(\displaystyle a=1, b= -4, c=12\); this is

\(\displaystyle b^{2} - 4ac = (-4)^{2}-4 \cdot 1 \cdot 12 = 16 -48 = -32\).

The discriminant being negative, there are no real solutions to this quadratic equation. Consequently, there are no points of intersection of the graphs of the two equations on the coordinate plane.

Substitute \(\displaystyle \frac{3}{x}\) for \(\displaystyle y\) in the second equation:

\(\displaystyle 4x-2y = -10\)

\(\displaystyle 4x-2 \left ( \frac{3}{x} \right )= -10\)

\(\displaystyle 4x- \frac{6}{x} = -10\)

\(\displaystyle 4x- \frac{6}{x} + 10 = -10+ 10\)

\(\displaystyle 4x- \frac{6}{x} + 10 = 0\)

\(\displaystyle \left (4x- \frac{6}{x} + 10 \right ) \cdot \frac{x}{2} = 0 \cdot \frac{x}{2}\)

\(\displaystyle 2x^{2}-3+5x = 0\)

\(\displaystyle 2x^{2}+5x -3= 0\)

This quadratic equation can be solved using the \(\displaystyle ac\) method; the integers with product \(\displaystyle 2 \cdot (-3) = -6\) and sum 5 are \(\displaystyle -1\) and 6, so continue as follows:

\(\displaystyle 2x^{2}-x+6x -3= 0\)

\(\displaystyle x (2x -1) +3(2x -1)= 0\)

\(\displaystyle (x+3 )(2x -1)= 0\)

Either \(\displaystyle x+3 = 0\), in which case \(\displaystyle x = -3\), or

\(\displaystyle 2x-1 = 0\), in which case

\(\displaystyle 2x = 1\)

\(\displaystyle x = 0.5\)

The desired \(\displaystyle y\)-coordinate is paired with the positive \(\displaystyle x\)-coordinate, so we substitute 0.5 for \(\displaystyle x\) in the first equation:

\(\displaystyle y = \frac{3}{x} = \frac{3}{0.5} = 6\)

Using the substitution method, set the values of \(\displaystyle y\) equal to each other.

\(\displaystyle \frac{2}{x} = -\frac{3}{x+1}\)

Multiply both sides by \(\displaystyle \frac{x(x+1)}{1}\):

\(\displaystyle \frac{2}{x} \cdot \frac{x(x+1)}{1}= -\frac{3}{x+1}\cdot \frac{x(x+1)}{1}\)

\(\displaystyle 2(x+1)= -3x\)

\(\displaystyle 2x+2= -3x\)

\(\displaystyle 2x+2 + 3x -2 = -3x + 3x -2\)

\(\displaystyle 5x = -2\)

\(\displaystyle 5x \div 5 = -2 \div 5\)

\(\displaystyle x = -0.4\)

Substitute in either equation:

\(\displaystyle y = \frac{2}{x} = \frac{2}{-0.4} = -5\)

The \(\displaystyle y\)-intercept of the graph of \(\displaystyle y = \frac{3}{2x-7}\)—the point at which it crosses the \(\displaystyle y\)-axis—is the point at which \(\displaystyle x = 0\), so substitute accordingly and solve for \(\displaystyle y\):

\(\displaystyle y = \frac{3}{2 \cdot 0-7}= \frac{3}{ 0-7}=- \frac{3}{ 7}\)

The \(\displaystyle y\)-intercept of this graph, and that of the line, is \(\displaystyle \left (0, - \frac{3}{ 7} \right )\). Since the slope is 4, the slope-intercept form of the equation of the line is

\(\displaystyle y = 4x-\frac{3}{7}\)

To put it in standard form:

\(\displaystyle y-y + \frac{3}{7} = 4x-\frac{3}{7} -y + \frac{3}{7}\)

\(\displaystyle \frac{3}{7} = 4x -y\)

\(\displaystyle \frac{3}{7} \cdot 7 = (4x -y) \cdot 7\)

\(\displaystyle 28x-7y=3\)

\(\displaystyle \angle B\) is a right angle and \(\displaystyle m \angle C = 30 ^{\circ }\), so 

\(\displaystyle m \angle A = 90 ^{\circ } - m \angle C = 90 ^{\circ } - 30 ^{\circ }= 60 ^{\circ }\),

making \(\displaystyle \bigtriangleup ABC\) a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, the length of the short leg \(\displaystyle \overline{AB}\) is half that of hypotenuse \(\displaystyle \overline{AC}\):

\(\displaystyle AB= \frac{1}{2}\cdot AC = \frac{1}{2}\cdot 30 = 15\)

and the length of long leg \(\displaystyle \overline{BC}\) is \(\displaystyle \sqrt {3}\) times that of \(\displaystyle \overline{AB}\):

\(\displaystyle BC = AB \cdot \sqrt{3}= 15 \cdot \sqrt{3}\)

Corresponding sides of congruent triangles are congruent, so \(\displaystyle DE = AB\); since \(\displaystyle AB=15\), it follows that \(\displaystyle DE = 15\).

Also, \(\displaystyle DE = AB = 15\), \(\displaystyle EF= BC = 15 \sqrt{3}\),  and \(\displaystyle DF = AC = 30\), so the perimeter of \(\displaystyle \bigtriangleup DEF\) is the sum of these, or 

\(\displaystyle DE+EF+DF = 15+15 \sqrt{3}+30=45+15 \sqrt{3}\).

Corresponding angles are congruent, so \(\displaystyle m \angle D = m \angle A\) and \(\displaystyle m \angle F = m \angle C\). By substitution, \(\displaystyle m \angle D = 60 ^{\circ }\) and \(\displaystyle m \angle F = 30 ^{\circ }\).

The false statement among the choices is that \(\displaystyle \angle F\)is a right angle.

\(\displaystyle f(x)=ax^2+bx+c\)

\(\displaystyle f(x)=-2x^{2}-4x+7\)

The \(\displaystyle x\)-coordinate of the vertex is \(\displaystyle x=-\frac{b}{2a}\), where \(\displaystyle a = -2, b=-4\).

\(\displaystyle x=-\frac{b}{2a} = -\frac{-4}{2 \left (-2 \right )} = -1\)

To get the \(\displaystyle y\)-coordinate, evaluate \(\displaystyle f (-1)\).

\(\displaystyle f(x)=-2x^{2}-4x+7\)

\(\displaystyle f(1)=-2 \left (-1 \right )^{2}-4\left (-1 \right )+7 =-2+4+7=9\)

The vertex is \(\displaystyle (-1,9)\).