does anyone knows how to do this?

(a) i) Vector BC = 3/2 a + 5b

ii) Vector AM = 15/4 a + 5/2 b

(b) Vector QP = -15/4 b where k = -15/4

Step-by-step explanation:

* Lets explain how to solve this problem

∵ ABCD is a trapezium

∵ AB // DC

∵ The vector AB = 3a

∵ Vector DC = 3/2 vector AB

∴ Vector DC = 3/2 × 3a = 9/2 a

∵ Vector AD = 5b

(a)

i) ∵ Vector BC = vector BA + vector AD + vector DC

∵ Vector AB = 3a , then vector BA = -3a

∵ Vector AD = 5b , vector DC = 9/2 a

∴ Vector BC = -3a + 5b + 9/2 a = (-3a + 9/2 a) + 5b

∴ Vector BC = 3/2 a + 5b

ii) ∵ Vector AM = vector AB + vector BM

∵ M is the mid-point of BC

∴ Vector BM = 1/2 vector BC

∵ Vector BC = 3/2 a + 5b

∴ Vector BM = 1/2(3/2 a + 5b) = (1/2 × 3/2) a + (1/2 × 5) b

∴ Vector BM = 3/4 a + 5/2 b

∴ Vector AB = 3a

∴ Vector AM = 3a + 3/4 a + 5/2 b = (3a + 3/4 a ) + 5/2 b

∴ Vector AM = 15/4 a + 5/2 b

(2)

∵ 7 DQ = 5 QC ⇒ divide both sides by 7

∴ DQ = 5/7 DC

∴ The line DC = 7 + 5 = 12 parts ⇒ DQ 5 parts and QC 7 parts

∵ DQ = 5/12 DC

∵ Vector DC = 9/2 a

∴ Vector DQ = 5/12 (9/2 a) = 45/24 a ⇒ divide up and down by 3

∴ Vector DQ = 15/8 a

∵ P is the mid point of AM

∴ Vector AP = 1/2(15/4 a + 5/2 b) = (1/2 × 15/4) a + (1/2 × 5/2) b

∴ Vector AP = 15/8 a + 5/4 b

∵ Vector QP = QD + DA + AP

∵ Vector DQ = 15/8 , then vector QD = -15/8 a

∵ Vector AD = 5b , then vector DA = -5b

∴ Vector QP = -15/8 a + -5b + 15/8 a + 5/4 b

∴ Vector QP = (-15/8 a + 15/8 a) + (-5b + 5/4 b)

∴ Vector QP = -15/4 b

∵ -15/4 is constant

∴ Vector QP = k b ⇒ proved

4096 bacteria

step-by-step explanation:

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )