CPhill

Alice, Bernice, Celine, Devi had a total of 156 stamps. Alice lost 22 stamps and Bernice bought 16 stamps. If Devi had 3 times the number of stamps she originally had, then all of them would have an equal number of stamps. How many stamps did Celine have at first?

Let w = Alice's original number of stamps, x =  Bernice's  original number of stamps, y = Celine's original number of stamps and z = Devi's original number of stamps

And we have this system :

w + x + y + z = 156    (1)

y = 3z   (2)

w - 22  = 3z   → w = 3z+ 22  (3)

x  + 16 = 3z  →  x = 3z - 16    (2)

And substituting (2), (3) and (4) into (1), we have :

[3z + 22] + [ 3z - 16] + 3z + z = 156      simplify

10z  + 6  = 156

10z = 150

z = 15

Celine had/has 3(15)  = 45 stamps 

Check

67 +  29 + 45  + 15  = 156

Let T be the total number of seats before the additional number were added

And let R be the revenue collected before the additonal seats were added

So we have this system

.75T(40) + .25T(75)  = R   (1)

.75T(40) + (1460 - .75T)(75) = R + 21900  (2)

Substituting (1) into (2), we have

.75T(40) + (1460 - .75T)(75) = .75T(40) + .25T(75) + 21900   simplify

 109500 - 56.25T  = 18.75T + 21900

75T =  87600

T = 1168

And after the seats were added, the number of VIP seats = (1460 -  .75*1168)  =   584

And the % of VIP seats after the seats were added = 584/1460  = 40%

By using the change-of-base method......for example....suppose we have

log₃ 7     ........the answer to this is given by:     log(7) / log(3)  ≈ 1.77

So......as long as a, b > 0    and a not equal to 1, we have

log_a b =     log(b) / log(a)

Thanks LambLamb.......

Remember that ......   2ⁿ / 2ⁿ   =  2^n-n  = 2⁰   but, as LambLamb points out...   2ⁿ/2ⁿ  = 1

So......   2⁰  = 1

Thanks, LambLamb for that good answer......!!!!

Here's another way to figure this out.....

Let T be the total travel time - in hours - for the slower bike....so T + 4  = the total travel time  - in hours- for the faster one....and we have Rate * Time = Distance.......so......twice the distance the slower one traveled = the distance the faster one traveled.......or, algebraically......

2[90T]  = 120[T + 4]  simplify

180T  = 120T + 480    subtract 120T from both sides

60T = 480     divide both sides by 60

T  = 8hrs

So....the slower bike travels .....   90kmh(8h) = 720km

And as LambLamb pointed out.......the faster one traveled 120km/hr (12h)  = 1440km

Here's the first one....alll the rest are just the same idea.......

The vertical position function is :   h(t)  = vsin(theta) t - (4.9)t^2     where v is the original velocity, theta is the launch angle, and h is the height at any time

So, taking the derivative of this and setting it to 0, we have

h'(t) = 20sin(15) - (9.8)t  = 0      and solving this for t, we have.....t = 0.528202 seconds......

And this is the time to reach the max ht

And that height = 20sin(15)(0.528202) - (4.9)(0.528202)^2   = 1.36709m

And the horizontal displacement (range) = v * t , where v is the intital velocity and t is the total time..... 20m/s * (2*0.528202)s  = 21.128m

y=-3(0)+2=  0 + 2 = 2

0.0622984679794434

9x+3-4x-2  =

5x + 1

210 = 2 x 3 x 35

210 = 2 x 3 x 7 x 5