Lines <em>a</em> and <em>b</em> are parallel, so lines <em>p</em>, <em>q</em>, and <em>t</em> are considered to be transversals. To solve this, you make use of the fact that alternate interior angles are equal, as are alternate exterior angles, as are corresponding angles. Of course any linear pair of angles is supplementary.
∠1 = 90° — corresponding angle to the right angle above it
∠2 = 68° — the sum of 22° and angles 1 and 2 is 180°
∠3 = 112° — supplementary to angle 2 (and the sum of 22° and 90°, opposite interior angles of the triangle)
∠4 = 112° — equal to angle 3
∠5 = 68° — equal to angle 2; supplementary to angle 4
∠6 = 56° — base angle of isosceles triangle with 68° at the apex; the complement of half that apex angle
∠7 = 124° — supplementary to the other base angle, which is equal to angle 6; also the sum of angles 5 and 6
∠8 = 124° — alternate interior angle with angle 7, hence its equal.
Answer:
a² + 2ab + b²
Explanation:
⇒ (a + b)²
⇒ (a + b)(a + b)
[apply distributive method: (a + b) (c + d) = ac + ad + bc + bd]
⇒ a² + ab + ab + b²
⇒ a² + 2ab + b²
The result should be 4 if it’s using PEMDAS
Round the numbers to 70 and 30, then multiply them together to get the product of 2100
