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3 years ago

If two parallel lines are cut by non-perpendicular transversal, which type of angles are NOT congruent? (geometry)

Mathematics

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

${\huge{\underbrace{\overbrace{\color{red}{Answer...}}}}} \\ \\$

$\small\boxed{\mathfrak{\underline{Interior \: angles \: are \: not \: congruent}}} \\ \\ {\huge{\underbrace{\overbrace{\color{purple}{hope \: it \: helps... | \: | }}}}} \\ \\$

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Given the function f(x) = (x-6)^2 + 1

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$=x^2-12x+37$ $\mathrm{The\: parabola\: params\: are\colon}$ $a=1,\: b=-12,\: c=37$ $x_v=-\frac{b}{2a}$ $x_v=-\frac{\left(-12\right)}{2\cdot\:1}$ $\mathrm{Simplify}$ $x_v=6$ $y_v=6^2-12\cdot\: 6+37$

Simplify:

$y_v=1$

$\mathrm{Therefore\: the\: parabola\: vertex\: is}$ $\mleft(6,\: 1\mright)$ $\mathrm{If}\: a$ $\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}$ $a=1$ $\mathrm{Minimum}\mleft(6,\: 1\mright)$ $\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$y=x^2-12x+37$ $\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}$ $a=1,\: b=-12$ $x=\frac{-\left(-12\right)}{2\cdot\:1}$ $\mathrm{Refine}$

Axis of simmetry : x=6

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3 years ago

￼ If two parallel lines are cut by non-perpendicular transversal, which type of angles are NOT congruent? (geometry)

If two parallel lines are cut by non-perpendicular transversal, which type of angles are NOT congruent? (geometry)