-="Math Clinic"=-

[Ebenezer Scrooge]

sana maging active ulit to

 

[rizza0806]

wala naman na ding mga nagtatanong...

 

[Ebenezer Scrooge]

pde ba ito gamitan ng trig substitution >> (a^1/2 - x^1/2)^4 ?

 

[coonaz]

Re: -="Math Clinic"=-

MOTION PROBLEMS
. Recall the Relationships among Distance,Rate, and Time in Physics
Distance = Rate x Time


1. Juan can Travel 80km while riding his calesa
in uniform speed in the same amount of time
he can run 30km. if he is 30kph slower while running
,how fast does his horse run at uniform speed?


pa solve mga boss thanks.

 

[Razielle Prima]

Re: -="Math Clinic"=-

80km/h r = 30km/h / (r - 30km/h)

> 80 (r - 30) = 80 r - 2400 = 30 r

> 80 r - 30 r = 50 r = 2400

> r = 2400/50 = 48

 

[machine1983]

Re: -="Math Clinic"=-

MOTION PROBLEMS
. Recall the Relationships among Distance,Rate, and Time in Physics
Distance = Rate x Time


1. Juan can Travel 80km while riding his calesa
in uniform speed in the same amount of time
he can run 30km. if he is 30kph slower while running
,how fast does his horse run at uniform speed?


pa solve mga boss thanks.

V = d / t; where: V = Rate, d = distance and t = time
Equation 1:
VH = 80 km / t ; where VH = Rate of the horse

Equation 2:
VJ = 30 km / t ; where VJ = Rate of Juan while running

Equation 3:
VJ + 30 kph = VH (found in the problem) or it can be written as VJ = VH - 30 kph (for substitution purposes)

Note that "t" in equations 1 and 2 are equal as stated in the problem.

Now, substitute Equation 3 into Equation 2 then make "t" as the subject of the equation. We have,
VH - 30 = 30 / t
t = 30 / (VH - 30)

Then make "t" as the subject of the equation in Equation 1. We have,
t = 80 / VH

Then equate: t = t
30 / (VH - 30) = 80 / VH
30VH = 80VH - 2400
50VH = 2400
VH = 48 kph (answer)

 

[schiniezelBRITANNIA]

salamat po