Answer:
14
Step-by-step explanation:
Substitute x = 2 into f(x)
f(2) = 7 × 2 = 14
If the parallel sides are the same length, then the figure must be a parallelogram. You can prove this by dividing the parallelogram into two triangles, and then using SAS (side angle side) to prove the triangles congruent, which leads to you showing the corresponding angles are the same measure, therefore the other set of sides must be parallel as well.
Or
If the non parallel sides are the same length, then you have an isosceles trapezoid. A trapezoid is any figure with exactly one pair of parallel sides. An isosceles trapezoid is one where the non-parallel sides are the same length. The non-parallel sides are sometimes considered the legs of the trapezoid (and the parallel sides are the bases).
Or
If you have two adjacent sides that are same length, and you have one set of parallel sides, then you could have a trapezoid (not isosceles but just a more generalized trapezoid)
Answer:
D
Step-by-step explanation:
Answer:
Step-by-step explanation:
1). m∠AEC = m∠AEB + m∠BEC
= 21° + 37°
= 58°
2). m∠BED = m∠BEC + m∠CED
= 37° + 44°
= 81°
3). m∠IKF = m∠IKH + m∠HKG + m∠GKF
= m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]
= 2∠IKH + m∠HKG
103° = 2∠IKH + 41°
2(∠IKH) = 103 - 41
m(∠IKH) = 31°
4). m∠AED = m∠AEB + m∠BEC + m∠CED
= 21° + 37° + 44°
= 102°
5). m∠JKG = 108°
m∠JKG = m∠JKI + m∠IKH + m∠HKG
108° = m∠JKI + 31° + 41°
m∠JKI = 108° - 72°
m∠JKI = 36°
6). m∠HKF = m∠GKF + m∠HKG
= m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]
= 31° + 41°
= 72°
7). m∠NQO = m∠MQN = 64°
8). m∠JKF = m∠JKI + m∠IKF
= 36° + 103°
= 139°
8). m∠MQO = 2(m∠NQO)
= 2(64)°
= 128°
9). m∠LQO = 156°
m∠LQM = m∠LQO - m∠MQO
= 156° - 128°
= 28°
10. m∠NQP = m∠NQO + m∠OQP
= 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]
= 64° + 28°
= 92°
