Need help solving a few small algebra problems.?

Question

I need detail on how to solve these problems and what the outcome will be, i did it for homework but i'm not sure i did it the right way.

1. Let w=width of a rectangle and 5w=the length. Translate the following into an algebraic equation: If you were to cut out four 1-meter squares (one from each corner), the remaining area is 200sq. meters.

2. Let r=the rate of the local train, and r+35 = the rate of the rapid train. Translate the following statement into an algaebraic equation: In four hours, the rapid travels twice as far as the local.

3. Several 6V and 12V batteries are arranged so that their individual voltages combine to be a poweer supply of 84V. how many of each type are present if the total number of batteries is 10.

4. A car travels 40kph for 2 hours along a certain route. Then a second car starts along the same route, traveling 60kph. When will the second car overtake the first.

5. In the study of forces on the wing of an aircraft, the equation M=R(L-x) is used. Solve for L in terms of other variables.

Thanks in advance! =]

Answer 1

1-The area of the remaining shape is the area of the first rectangle minus for subtracted corners, algebratically 5w*w-4*1=200--->w=6.4(app.)

2-We know that according to the definition of speed assuming constant speed d=V*t(d=distance travelled, V=Velocity, t=time of travel)...so assuming same start point:  (r+35)*4=2*r*4--->r=35(exactly)

3-We know that ideal batteries can not be paralelled so that all of them must be connected in series and in series connection the voltages sum up. Consider n the number of 6V batteries and m the number of 12V batteries. The equation is telling us that 6n+12m=84. The aux. equation is telling us that n+m=10. Eliminating m from the aux. eq. and combination with the main eq. we will lead us to 6*n+12*(10-n)=84--->n=6 so m=4

4-Same as second problem and assuming same start point 2*40+t*40=t*60--->t=4(4 hours after the second cars starts traveling it will catch the first)

5-This is a first order equation that was first solved by mohammad khawrazmi( http://en.wikipedia.org/wiki/Muhammad_ib... )...Well today solving it doesnt take one even a blink, L=(M/R)+x

Answer 2

1.(w-2)(5w-2) = 200

2.By rate do you mean speed? then it is 4(r+35) = 2(4r).

3.n(6)+(10-n)(12) = 84. But it depends on parallel and series connections. n = no of 6V batteries.

4.It depends on where the second car starts from on the same route.

5.L = x+(M/R)

Answer 3

1.  5w X w does not equal 200; but it does equal 200+4 before we cut out the corners

  Each corner has an area of 1

Equation:  5w^2 = 204

The answer the first guy gives for this question is slightly incorrect, he's not measuring the rectangles in between the cut corners.

2.  Set two equations equal to each other so that both trains algebraically travel the same distance.

  2(4r) = [4(r+35)]

Notice the abritrary 2 is multplied to the slower train so the distance traveled by each is equal for speed "r".

3.  You need two equations to solve for combinations of batteries that add up to 84V using exacty 10 batteries

      6x + 12y = 84

      x + y =10

to solve this rearrange such that y = 10-x and plug in for y in the top equation

6x + 120 - 12x = 84

x=6

y=4

check 6(6) + 12(4) = 84 volts, always check like this

4.  When fast car starts the other guy is already 80 kilomiles away

40t + 80 = 60t

solving for t, t=4 hours.  Notice this is 4 hours after the fast guy starts and is dead even with the first.  After 6 hours then the fast car overtakes the first.

5. M=R(lL-x)

(M/R)=L-x

(M/R) + x = L