Just used logic: Car 1 pulls away by 12 km for every 1 hour it travels. After 6 hours Car 1 will be 72 (6x12) km ahead. Car 2 covers that 72km in the remaining 1 hour, so Car 2's speed is 72 kph. So, Car 1's speed must be 84 (72+12) kph.

If time (t=hr) is the common unit for each trip duration, and the average difference of speed is also t=hr, can this problem be solved as a multiplication of fractions so that (12km*6hr=72km/hr), (12km*7hr=84km/hr)?

MR. TabletClass Math , thank you for another classical word problem in Algebra. Word Problems are challenging and fun to solve.

Just completed! Therefore we have S + 12 = d/6 and S = d/7; 6S + 72 = d and 7S = d; 6S + 72 = 7S; S = 72; Therefore the other speed must be 84 or 60. 84 fits best in substituition with time time value for s = d/t!

A (speed of car A) and B (speed of car b). A-12=B; 7B=6A (B=6/7A); A-12=6/7A;1/7A=12;A=84 kp; B=72kph; distance from gas station to Tampa is 504 km

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Xmph * 6 hr = d mi (X-12)mph * 7 hr = d mi 6x = 7x- 84 X = 84, so: 84mph, 72mph Check: 6hrs*84mph = 504 mi 7hrs*72mph= 504 mi

D = Unknown distance (D/6) - (D/7) = 12 (7D/42) - (6D/42) = 12 D/42 = 12 D = 504 Now you have everything you need :)

Professor for problems measuring the difference in speed if you need to quickly get the answer could you multiply the difference and the rate of the slow car and multiply to get the answer? for example in the problem the difference 12 x 7 the slow car equals 84

Great Work Bro. Trying to make the things easy for the students. I also do the same. Keep it up

@shanemckenzie8681

2 жыл бұрын

You guys are awesome for taking the extra effort ON TOP of your profession.. THANK YOU!

of course you can then take the rate*time and figure out that they each drove 504K, means they probably started somewhere in central Georgia ;)

my calculations are all correct John! x=72, x+12=84!

one way to solve this problem is 7 / 6 = ( v + 6 ) / ( v - 6 ) and then you get v = 78. substitute back and get v - 6 = 72 km / h and v + 6 = 84 km / h

ok... there's two equations in two unknowns... step 1 is find them. step 2 is to find s.1 and s.2 eq.1: s.1 - s.2 = 12 km/hr eq.2: ? interum finding, time to destination: t.1 = 6hr t.2 = 7hr develop relations between time (t) distance (d) speed (s) d = s × t d.1 = d.2 d.1 = s.1 × t.1 d.2 = s.2 × t.2 so s.1 × t.1 = s.2 × t.2 t.1 = 6hr t.2 = 7hr s.1×6hr = s.2×7hr from eq.1: s.1-s.2= 12km/hr s.1 = 12km/hr +s.2 (12km/hr+s.2)6h=s.2×7h 72km + (s.2)(6hr)=(s.2)(7hr) 72km= (7hr)(s.2)-(6hr)(s.2) 72km=(1hr)(s.2) s.2=72km/hr eq.1: s.1 - s.2 = 12 km/hr s.1=12km/hr+s.2 s.1=12km/hr+72km/hr =84km/hr VERIFY does d.1 = d.2 d.1 = s.1 × t.1 = 84km/hr×6hr = 504km d.2 = s.2 × t.2 = 72km/hr × 7hr = 504km d.1==❤ d.2 ✔️✔️✔️

I need a program that will prep me for math I need to take engineering courses..specifically electrical

This guy has verbal runaway!

Turning a written question like this one into an equation certainly does help to keep the old brain cells active. Let's call the speed of the slower car S .. and the faster car S + 12 The faster car took 6 hours .. and the slower car 6 + 1 = 7 So, I think (HOPE !) that the equation is 6 (S + 12) = 7S 6S + 72 = 7S 72 = S The car speeds are 72mph and 84 mph

math starts at 6:32

849 = 749

Find my mistake Us:speed of slow car:. U=distance/time=d/time Uf:speed of fast car Uf-Us=12km/hr= d/6hr-d/7hr=12km/hr 7d-6d/42=12 d=524km Uf=524/6=87.333km/hr Us=524/7=74.858 Please show me where I went wrong Thank you

@quench100

2 жыл бұрын

d=504

@user-ri6rn7ti5h

9 ай бұрын

849 car 1 749 car 2

@user-ri6rn7ti5h

9 ай бұрын

=3084 =3.5 98

Yikes.

I hate word problems. And the presentation suffers due to words getting in the way. First you say the "rates" of these cars are respectively 72 and 84 kph (You do not say these quantities represent an average of different velocities for each car; you do not show any such work.) Though something else may be meant. We still never determine the "average" rate of each car. But "78" is the average rate of both cars (156/2) simply stated. And this quantity never makes an appearance. If 72 and 84 constitute answers -- then what can the meaning of an "average" rate for each car signify? The answer is not here. 72 and 84 cannot be constant velocities and averages simultaneously. If they are the averages, since they can be interpreted quite as well as constants, then we need to know what minimum and maximum values of velocities they correspond to. 66 and 78, for example, easily represent lower/upper limits for an average of 72 on the slow side. 78 and 90 work for an average of 84 on the fast side. Thus an average difference of 12 kph. But this seems, on evidence, too much to assume. Still, the math is self evident, for what it represents.

Need a larger screen on this channel. Gee, you looking at it a sentence at a time. Not good.