Question

quiz #2
show all work and include units to receive full credit! circle your answers.

1. find the product and simplify. leave the answer in standard form.

(9x - 8)(4x + 9) =

1. solve for the missing angles: m∠a, m∠b, and m∠c.

Explanation:

Step1: Expand using FOIL method

$(9x - 8)(4x + 9)=9x\times4x+9x\times9-8\times4x - 8\times9$
$=36x^{2}+81x-32x - 72$

Step2: Combine like - terms

$36x^{2}+(81x - 32x)-72=36x^{2}+49x - 72$

Step1: Use the angle - sum property of a quadrilateral

The sum of the interior angles of a quadrilateral is $360^{\circ}$.
We know three angles: $115^{\circ}$, $70^{\circ}$, and $7^{\circ}$. Let the fourth angle be $y$.
$y=360-(115 + 70+7)=360 - 192=168^{\circ}$

Step2: Use linear - pair and angle - sum in triangles

For angle $A$:
We know that the angle adjacent to $115^{\circ}$ in the triangle with $A$ is $180 - 115=65^{\circ}$
In the triangle with $A$, using the angle - sum property of a triangle ($180^{\circ}$), and knowing one angle is $y = 168^{\circ}$ and the other is $65^{\circ}$
$A=180-(168 + 65)=180 - 233+180=17^{\circ}$
For angle $B$:
We know that the angle adjacent to $70^{\circ}$ in the triangle with $B$ is $180 - 70 = 110^{\circ}$
In the triangle with $B$, using the angle - sum property of a triangle. Let's assume the third angle in this triangle is $z$.
We know that the sum of angles around a point is $360^{\circ}$. Let's consider the non - interior angles of the quadrilateral.
The angle adjacent to $A$ in the non - quadrilateral part is $180 - A$.
In the triangle with $B$, we have:
The sum of angles in the triangle with $B$ gives us:
$B = 180-(110+17)=53^{\circ}$
For angle $C$:
$C = 7^{\circ}$

Answer:

$36x^{2}+49x - 72$