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.
Rational between 5.2 and 5.5
5.2 × 10/10 = 52/10
5.5 × 10/10 = 55/10
rational no. b/w them is 53/10 = 5.3
rational because it is in p/q form where q is not zero and p and q are integers.
irrational.no. b/w 5.2 and 5.5
5.20102030405060708090.......
irrational.no because it is non terminating and non recurring
Answer:
The triangles are congruent by Angle Side Angle congruency statement and the reason is below.
Step-by-step explanation:
Given:
∠ VUW ≅ ∠ XYW
VW ≅ YW
To Prove:
Δ VUW ≅ Δ XYW
Proof:
In Δ VUW and Δ XYW
∠ VUW ≅ ∠ XYW ……….{Given}
VW ≅ YW ............{Given}
∠ VWU ≅ ∠ XWY …………..{Vertically Opposite Angles are equal}
Δ VUW ≅ Δ XYW .….{ By Angle-Side-Angle congruence test}
......Proved
Answer:
The answer to your question are letters A, C and D.
Step-by-step explanation:
To factor this polynomial we need to look for the common factor of the three terms.
To find it, get the greatest common factor
140 28 14 2
70 14 7 2
35 7 7 5
7 7 7 7
1 1 1
Greatest common factor = 2 x 7 = 14
Factor the polynomial
140c + 28 - 14a = 14 ( 10c + 2 - a)
Factor only 7 7(20c + 4 - 2a)
Factor only 2 2 (70c + 14 - 7a)
