At noon, ship a is 100km due east of ship b. ship a is sailing west at 12km per hour, and ship b is sailing south at 10km per hour. at what time will the ships be nearest to each other, and what will this distance be ?

at 4:55 pm

About 64 km

Step-by-step explanation:

Let t be time in hours taken by ship A to   point  B and ship B takes time t hours to point A

Speed of ship A=12 km/h

Speed of ship B=10 km/h

Distance=

Distance travel by ship A=12 t

Distance travel by ship B =10 t

OA=10 t

OC=100 km

OB=OC-BC=100-12 t

Using pythagorous theorem

The distance between two ships is minimum

Differentiate w.r.t time

Again differentiate w.r.t

Hence, the distance is minimum

The time will be 4:55 pm

Substitute t=4.92

Then AB=

AB=64.01 Km

AB=64 km

Hence, the distance between two ships =64 km (about)

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116