CPhill

Two points on the graph are  (-2, -2)  and (2, -5)

And the slope is given by :

[ - 5 - (-2)] / [ 2 - (-2) ]  =    -3 / 4

Because the chain moves at the same speed over both sprockets......the rear sprocket must make 7/4 revolutions for every one revolution made by the front sprocket

So.....when the rider moves the front sprocket through 190 revs/min......the rear sprocket moves through 190 (7/4)  = 332.5 revs/min......and the back wheel will rotate at the same rate

So....the linear speed of the bicycle in inches per minute  =

2pi * 9 * 332.5   = about 18802.43  in/min

The speed in mph =

[2pi *9 * 332.5]inches per minute * 60 minutes in 1 hr / [63360 in/ mile]   = about 17.27 mph

log₄ 64-log₄(1/16)   .....  this can be written as

log₄ 64  - log₄ 4^-2   =

3  - (-2) log₄ 4  =

3 - (-2) * 1  =

3 - (-2)  =

5

e^(ln 5)   = 5

18,200 (.91)^t    gives the number of jobs after t years.....so....when t = 11, we have

18,200 (.91) ^11  = about 6449 jobs   [ 6400 to the nearest hundered]

x(t)   = -5 + 5 sin (2 pi * t / 13)

y(t)  = 3 + 5cos (2pi * t / 13)

log6=log(4x - x^2)   we need to solve this :

6 = 4x - x^2      rearrange as

x^2 - 4x - 6  = 0  .... this will not factor......the [potential] solutions are :

x =  2 + √10  ≈ 5.1623  ... or  x = 2 - √10 ≈ -1.1623

However......both [potential] solutions produce a negative log on the right......so.....there are no real solutions to this

I think this is supposed to be :

13-3^(3x)= -35  rearrange as

48 = 3^(3x)     take the log of both sides

log 48= log3^(3x)    and we can write

log48  = (3x)*log 3      divide both sides by 3log3

[log 48] / [ 3 log 3]  = x ≈  1.17

Ine^A + log10^B - e^InB - A    =

A*lne + B*log10 - B - A  =

A + B - B  - A   =  

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