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No prob   !!!

Second one.....

Remember that the inverse reverses  the coordinates

So....on the inverse graph we should have the points

(-6.4, 3)    (-2.6, 2)    (-1.2, 1)    (-1, 0)   ( -0.8, -1)   ( 0.6, -2)    (4.4, - 3)

Note that the graph should apear  to "fall"  from left to right  and the point (-1,0)  is on the graph

So    "A"  appears to be correct

ohhh okay, i see where  I made my mistake, thank you for your resonse! I appreciate it.

First one

We are using these properties here  (these work the same with Ln)  :

Log (ab)  = Log (a) + Log (b) 

Log (a/b)  = log a  - Log b 

And

b log a  = log a^b

^So......

log₃ (cd⁴)  =  log₃ (c)  + log₃ d⁴   =  log 3 (c)  + 4log 3 (d)    FALSE

(3/4)  ( Ln a + Ln b)   = (3/4)Ln ( ab)  = Ln (ab)^3/4  = Ln (a³b³)^(1/4)  = Ln ⁴√(a³b³)   TRUE

3 Ln e   - 2 Ln f    =  Ln e³  -  Ln f²   =  Ln ( e³ / f²)    TRUE

a  =  sqrt  (8^2 + 5^2 ]   =  sqrt  [ 64 + 25 ]  =  sqrt [ 89]  ≈ 9.4  cm

b  =  sqrt [ 17^2 - 12^2 ]  =   sqrt  [ 289 - 144]  =  sqrt [ 145 ]   ≈  12  cm

BC = 15.

Distance  / Rate   =  Time

Let  spped of plane in still air   = P

Let speed of  wind  = W

On  the trip out   the effective speed of the plane =  (P - W)    .....the wind is slowing the plane down

On the return trip  the feective  speed of the plane = (P + W)......the wind is speeding the plane up

On  the   trip out.....we have

180 / ( P - W)  =  2

180  = 2(P - W)

180 / 2  = P - W

90  = P - W

90 + W  = P         (1)

On the trip back,   (1hr 15 min  = 1.25 hrs)  we have that

180 / (P + W) =  1.25

180 / 1.25  = P + W

144  = P + W         sub (1)  into this  for P

144 = (90+ W)  + W    subtract  90 from both sides

54  =  2W    divide both sides by 2

27 = W  =  wind speed  (in mph)

And using (1)....90 + W  = P    ....so.......90 + 27 =   117   = speed of plane in still air (in mph)

His speed to Bismarck =180 / 2 =90 mph

His speed going back  =180 / 1.25 = 144 mph

His speed in still air =[90 + 144] / 2 =117 mph

Wind speed =[144 - 90] /2 = 27 mph

First thing  is that there are  1000mL   to each litre

So....we have this equation

(N ) (.05)  + (1000 - N) (.20)  =   1000(.14)        simplify

.05N  +  200 - .20N  =  140

-.15N  + 200  = 140    subtract  200 from both sides

-.15N  = -60      divide both sides by   -.15

N = 400 =  mL  of 5%  solution

1000 - N  =  1000 - 400  =   600mL  of 20% solution 

Similar to the last problem.....we have this equation

(N)(6)   +  ( 110 - N)(10)   = 820

6N + 1100 - 10N  =  820

-4N + 1100  = 820           subtract  1100 from both  sides

-4N  = -280      divide both sides by  -4

N  = 70  =  number of boxes of standards

110 - N   = 110  - 70  =  40  = number of boxes of deluxes

Regular gas =R galons
Premium gas=310 - R galons

2.30R + 2.85[310 - R] =746, solve for R
R = 250 galons sold.
Premium =310 - 250 =60 galons sold

Let the number  of dimes  =   N

So.....the number of quarters   must just be  14  - N

And

Number of dimes * value each  + Number of quarters * Value each  =  Total Change

(N) (.10)  + (14 - N) (.25)  =  2.00           simplify

.10N  + 3.5 - .25N  =  2.00

-.15N  + 3.5  =  3      subtract 3.5 from both sides

-.15N  = -1.5          divide both sides by  -.15

N =  10  =   number of dimes

14 - N  = 14   - 10   =   4   = number of quarters

Why are you wasting our time fucking around on here?

Go watch p o r n. Search for the strap-on action vids.

91620    74848     70363    06857     52569    06161    86277    49814

The numbers  are

207, 748 , 363

Find $a+b+c$ if the graph of the equation $y=ax^2+bx+c$ is a parabola with vertex $(5,3)$, vertical axis of symmetry, and contains the point $(2,0)$.

We have the vertex form

y  = a(x - h)^2  +  k            where  (h, k)  =  ( 5, 3)

And since the point (2,0)  is on the graph, we can slove for "a"  thusly

0  =  a(2 - 5)^2  + 3      subract 3 from both sides    

-3  = a(-3)^2

-3  = 9a           divide both sides by  9

(-1/3)  =  a

So  we have

y = (-1/3) (x - 5)^2  +  3

y = (-1/3) [ x^2 - 10x + 25)  + 3

y =(-1/3)x^2  + (10/3)x - 25/3  +  9/3

y  = (-1/3)x^2 + (10/3)x -  16/3

So

a +  b  +  c   =

(-1/3)  + (10/3) +  (-16/3)  =

- 7 / 3

Area of right triangle  =  (1/2)(BC ) (BA)       (1)

Two white areas  that  are between the   gray areas  and triangle ABC  =  area of semi-circles  - area of  ABC  =

(1/2)  [ pi (AC/2)^2 -  (BC)(BA)  ]  =   (1/2) [ pi ( BC^2 + BA^2)/4  - (BC)(BA)   ]   (2)      

Area   of   grey semi-circles  =  (1/2)pi [  (BC/2)^2  + ( BA/2)^2  ]   -  (2)   =

(1/2) pi  [ (BC^2 + BA^2)/4 ]   -  (1/2)  [ pi (BC^2  + BA^2)/4 - (BC)(BA) ]   =

-(1/2) [ - (BC)(BA)  ]   =

(1/2)(BC)(BA)   which  =  (1)

So....true  !!!

Who is this person?????!!!!!!!

submitting an online survey

The largest possible difference  occurs   when

a b   -  c d   =

98   -  21   =

77

Or.....if we are multiplying

a* b  - c * d  =

9 * 8  -  2 * 1   =

72  -  2   =

70