Rom

\(\Large \dfrac{a^{2n-1}\times b^3 \times c^{1-n}}{a^{n-3}\times b^{2-n}\times c^{2-2n}} = \\ \\ \Large a^{(2n-1)-(n-3))}\times b^{3-(2-n)}\times c^{(1-n)-(2-2n)} = \\ \\ \Large a^{n+2}\times b^{n+1}\times c^{n-1}\)

ok two steps to this

first we find the slope of the line passing through the two points

\(m = \dfrac{4-1}{4-6} = -\dfrac{3}{2}\)

now we want a line perpendicular to this so we have a slope of

\(m_{\perp} = -\dfrac{1}{m} = -\left(\dfrac{1}{-\frac 3 2}\right) = \dfrac 2 3\)

now using point slope formula with the point (-2,6) we have

\((y-6) = \dfrac 2 3 (x - (-2)) \\ y = \dfrac 2 3 x + \dfrac{22}{3} \)

parallel means same slope

if line has slope \(m\), then perpendicular line has slope \(m_{\perp} = - \dfrac{1}{m}\)

really....

\(A = s^2 \\ 225 = s^2 \\ s = \pm \sqrt{225} \\ s = \pm 15 \\ \text{negative lengths don't make sense}\\ s = 15\)

Thinking it has to be (c) but you're the one in the class that's teaching this stuff.

i.e. \(\dfrac{10\%}{12}=0.833\%\)

\(180^\circ \text{ is the sum of the interior angles of a triangle}\\ x = 180-42=73 = 65\\ \text{the sum of supplementary angles is }180^\circ \\ y = 180 - 131 = 49 \\ \text{Now z is supplementary to the unlabeled angle that is }180-x-y \\ \text{thus }z = 180 - (180-x-y) = x+y \\ z = 65 + 49 = 114^\circ\)

Presumably solutions means solutions to f(x) = 0

In this case there are 3 unique solutions, you can read them right off as

x = 3, x = -2, x = 1

in (2)

a, b, and c are the values for which f(x) = 0, i.e. crosses the x axis

a) Yes, as long as a and b are allowed to be complex numbers.  Consider the simple parabola

\(y = x^2+1 \\ \text{with complex roots } \pm i \\ \text{this can be written as }\\ (x-i)(x+i) = x^2 +i x - i x +(i)(-i) = x^2+1\)

b) No.  Consider a general cubic polynomial

\(y = a x^3 + b x^2 + c x + d\)

\(\text{as }x \to -\infty, \text{the }x^3 \text{ term dominates everything and }y \to -\infty \\ \text{as }x \to +\infty, \text{ likewise the }x^3 \text{ term dominates and }y \to +\infty \\ \text{somewhere along the way it's got to cross the x axis and at that point }\\ y=0 \text{ at a real value of x}\)