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Glad to save a life when I can.....

No prob....!!!!

Thanks CPhill!! You're a lifesaver haha.

OK....this IS a little tough

A  "fraction of 12" can be represented  as  (1/x)*12 

Taking this away from  17   is      17  -  (1/x)*12

So we have

[ 17 - (1/x)*12 ]    -  (1/3)*17    =   6

thank you Cphill

https://web2.0calc.com/questions/circle-is-the-incircle-of-triangle-abc-and-is-also_1

This has the answer!!!!!

The first is definitely arithmetic

The common difference between terms   is  4

The sequence can be expressed as

a_n+1 =  a_n + 4       where   a₁ = - 8

I'm not used to finding an graph for these, but here's a possibilty

y = -8 + 4(x - !)   

y = -8 + 4x - 4

y = 4x - 12       

[ Where x is an integer...and  x = 1  gives the 1st term, -8, x = 2 gives the 2nd term, -4.....etc .]

Here's the graph :    https://www.desmos.com/calculator/pewqtw4vmg

The second is not arithmetic....the difference between 2 and 4  is 2......the difference between 8 and 4 is 4.......thus.....the common difference between terms is not constant as required in an arithmetic series

You have to divide both sides by 23 and then calc a logaritm (log) with base 0.94 of the right side so its gonna be x=log0.94(15/23) web2.0calc doesn't have an option for any base of logarythm so you can just calc ln(15/23)/ln(0.94) and the answer won't change. By the way answer is around 6.908.

Thank you very much for explaining this difficult one for me. Substitution is much easier to work with.

No prob...!!!!

Thank you for simplifying this problem for me!

Thank you for helping me and explaining the process! With these log based problems, sometimes I have a hard time remembering.

\(23(0.94)^x = 15\\ (0.94)^x = \dfrac{15}{23}\\ x\log(0.94) = \log\left(\dfrac{15}{23}\right) = \log(15)-\log(23)\\ x = \dfrac{\log(15)-\log(23)}{\log(0.94)}\)

\(\text{Let }\Phi(x) \text{ be the CDF of the standard normal distribution}\\ P[n \leq N] = \Phi\left(\dfrac{N-\mu}{\sigma}\right)\\ P[N < n] = 1-P[n \leq N] \\ P[N_l \leq n \leq N_h] = \Phi\left(\dfrac{N_h-\mu}{\sigma}\right) - \Phi\left(\dfrac{N_l-\mu}{\sigma}\right)\)

now just dump your numbers in

Formula

New Point = [ old point - dilation center ] * scale factor + dilation center

So

For   (-1, 1) 

[ (-1, 1) - (-7, -2) ] * 2  + (-7, -2)  =  [ 6, 3] *2 + (-7, -2)  = [ 12, 6] + (-7, -2) = (5, 4)

For (-3, -2)

[  (-3, -2) - (-7, -2) ] *2  + (-7, -2) = [ 4, 0]*2 + (-7, -2) = [ 8, 0 ] + (-7, -2) = (1, -2)

For (-4, 3)

[ (-4,3) - (-7 , -2) ]*2 + (-7, -2)  = [ 3, 5]*2 + (-7, -2) = [ 6, 10] + (-7, -2) = (-1, 8)

For (-6, 2) 

[ (-6, 2) - (-7, -2) ] *2 + (-7, -2) = [ 1, 4]*2 + (-7, -2) = [ 2, 8] + (-7, -2) = (-5, 6)

Graph the points in red and connect them to form a quadrilateral

Yep....a little tough!!!.....the idea is to isolate x

2^(2x)   = 5^(x - 1)         take the log of both sides

log 2^(2x)   =  log 5^(x - 1)       using a log property, we have

2x* log 2  = (x - 1) log 5       simplify

(2log 2) * x   =  xlog 5  - log 5          rearrange as

log 5  =   x log 5 - (2log2) x          factor out x on the right

log 5  =  x [ log 5 - 2log2 ]         divide  both sides by  [ log 5 - 2log 2]

log 5

___________   =   x           {second answer }

log 5 - 2 log 2

Note Phoenix that I used log  instead of Ln.....just substitute.....the answers will be the same, numerically !!!!!!

2^t = 38       take the log of both sides

log 2^t  = log 38        and we can write

t * log 2  =  log 38      divide both sides by  log 2

t =   log (38) / log(2)   ≈  5.2479

Note, Phoenix.....this implies that any exponential equation in the form  a^t = b    will have the solution of   log b / log a   .....!!!!

Hey, shrek....how was the play ????

10^x  = 97

Take the log of both sides

log 10^x =  log 97       and we can write

x * log 10  =  log 97

[ Note that log 10 = 1 ]

x * 1   =  log 97

x = log 97

I'm not totally comfortable with sig figs, but I believe that the answer can only be expressed as accurately as least number of sig figs  of any data used in the computations.....thus ....the answer should be rounded to 2 sig figs

So  

187 / 28   =  6.67   ⇒  6.7 km / L     [rounded to 2 sig figs ]

Initial Payment = FV [  (r-g) / ( 1+r)^n - (1+g)^n) ]      r = .05  g = .03  n = 40   FV = 377795.09

                         = 1999.999 ~~~ $2000.

Karen also took my kids away after she filed the divorce

Sorry Jenny.....yes... it should be

log₃ (x)  + 1

I'll make that correction....thx for pointing that out....it's important  !!!

what about the (  ) do we need those

The idea, at first, is to get x by itself

Let's write this

y = 3^(x - 1)       take the log of both sides

log y = log 3^(x - 1)     and we can write

log y = (x - 1) log 3      divide both sides by log 3

[ log y  ] / [log 3]  = x - 1      add 1 to both sides

[ log y ] / [ log 3]  + 1  =  x       now,  "swap" x and y

[log x] / [log 3] + 1  =   y           

By the change of base theorem,  [ log x ] / [ log 3]  =  log₃ (x)

So we have

log₃ (x)  + 1  =  y  =  f^-1(x)         and this is the inverse