Let jk be the median of lmba. find jk, m∠a and m∠alm

JK = 83 , m∠A = 70° , m∠ALM = 110°

Step-by-step explanation:

* Lets explain how to solve the problem

∵ ABCD is a trapezoid

∴ DC // AB

∴ m∠D + m∠A = 180° ⇒ interior supplementary angles

∵ m∠D = 110°

∴ 110° + m∠A = 180° ⇒ subtract 110° from both sides

∴ m∠A = 70°

∵ L is the midpoint of AD, and M is the midpoint of BC

∴ LM is the median of trapezoid ABCD

∴ LM // AB and DC

∴ m∠D = m∠ALM ⇒ corresponding angles

∵ m∠D = 110°

∴ m∠ ALM = 110°

- The length of the median is half the sum of the lengths of the two

 parallel bases

∴ LM = 1/2 (AB + DC)

∵ AB = 96 units and DC = 44 units

∴ LM = 1/2 (96 + 44) = 1/2 (140) = 70 units

- In the quadrilateral ABML

∵ AB // LM

∵ AL ≠ BM

∴ ABML is a trapezoid

∵ JK is its median

∴ JK = 1/2 (AB + LM)

∵ AB = 96 units ⇒ given

∵ LM = 70 units ⇒ proved

∴ JK = 1/2 (96 + 70) = 1/2 (166) = 83

∴ JK = 83 units

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

n

step-by-step explanation: