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I think the answer is D because the net gain is $37,500, which is a positive amount. 

Thanks so much, CPhill! I fully understand the solution now! 

Solution to second question is \(a = -30.5\).

I am pretty sure the answer is C. It is true that the strength training program does not reduce or increase the number of muscular injuries that occur during an athletic event. Since the injuries stay the same, the numbers do not reduce, so the answer is C.

But since the average of the 5 athletic events was taken, there could be more injuries in one event, so the answer D could also be considered, but I think the correct answer would be C. 

f(x)   =  ax + b    for f(x), write y

y = ax + b      subtract b rom both sides

[y - b]  =  ax   divide both sides by a

[ y - b] / a  =  x      "swap' x and y

[x - b] / a  = y  

(1/a)x - b/a  = y  =   f^-1(x)

So g(x)  = f^-1(x) - 3  =   (1/a)x - b/a - 3  =  5x - 4

This implies that

(1/a) = 5       and    -b/a - 3  = 4

So

a = (1/5)      and 

-b / (1/5) - 3  =  -4     add 3 to both sides

-b / (1/5)  =  -1

-b =  -1 (1/5)

-b = -1/5

b = 1/5

So

5a + 5b  =

5(1/5) + 5(1/5)  =    1 + 1   =   2

I think I've seen this question a few days ago...I am sorry though, I do not know how to solve this because I do not fully understand the problem. 

umm I did r^2 = 20 then r = 10...?

What scatterplot did you generate???

We have a horizontal line.....therefore....the slope   = 0

This can be confirmed by the "0"  in the linear function

y = 0r + 40

oh right ok thanks

If r = 0.....there is no correlation

A

10             M                                      cos ABM

B        24             C

Since BM is an angle bisector of  right angle ABC....then angle ABM  = (1/2)(90°) = 45°

So....the cos ABM  = cos (45°)   =  1 / √2

If we have an isosceles right triangle....the legs will be of equal length

AC is one leg and AB  is the other

And they both equal 6cm

And they form the base and height of the triangle

And the area of a right triangle  =

  [ Base * Height} / 2  = [ Product of the leg lengths ]  / 2   =

[6 * 6 ] / 2  =

36 / 2  =

18 cm^2

y = $1,200 x 1.02^x

I think most of your troll posts are hilarious.  What kind of pun would pissoff your cat? Whatever it was, funny or not, it’s now enigmatic!! 

P               Q

6      X

S      8         R

This is similar to your other problem

Triangle PSR  forms a  6-8-10  Pythagorean Right triangle with PR   = 10

So....PX  =  SX  =  (1/2) (10) =  5

Using the Law of Cosines

PS^2  = PX^2 + SX^2  - 2(PX * SX)cosPXS

6^2 = 5^2 + 5^2  - 2 (5 * 5)cosPXS

36 - 2(5)^2

_________  =  cos PXS

 -2(5)^2

36  - 50

______    =   cos PXS

  -50

50 - 36                           14            7

_____   =  cos PXS  =   ___ =     ___

   50                               50           25

Since  PXS is acute.....then its sine will be positive  and we can find it as

sin PXS  = √ [ 1 - cos^2 PXS]  =  √ [ 1 -  (7/25)^2 ]  = √ [ 1 - 49/625]  =  √ [625 - 49]/ 25  =

√576/ 25   =    24 / 25

P                    Q

10       A

S       24         R

We can use the Pythagorean Theorem to   find PR  =

√ [ SR^2 + PS^2]   = √ [ 576 + 100]  = √676  =  26

And  (1/2) of this  = PA  = SA  =  13

So.....using the Law of Cosines

PS^2  = PA^2 + SA^2  - 2(PA * SA) cos PAS

10^2 = 13^2 + 13^2 - 2(13 * 13) cos PAS

100 - 2(13)^2

___________  =   cos PAS

  -2 (13^2)

2(13)^2  - 100

___________   =   cos PAS

   2(13)^2

338 - 100

________    =    cos PAS

  338 

238                                119

___    =   cos PAS =     ____

338                                169

There must be some error in logic here:

Let r be the common ratio

Let the first term  = a

So.....the third term is   ar^2

And the sixth term is  ar^2

So.....we have this ratio

ar^2          (4/3)   

____  =    _______

ar^5          (-32/27)

r^2           (4/3)

___  =     ______       and we can invert these fractions and write

r^5          (-32/27)

r^3   =  (-32/27)  / (4/3)

r^3  =  (-32/27) * ( 3/4)   =  (-8) / (9)

r =  (-8/9)^(1/3)  =   -2 / (9)^(1/3)   =   - 2 / (3^2)^(1/3)  =  - 2 / (3)^(2/3)

So....to find the second term, we have

(2nd term) * r  =  (3rd term)          divide both sides by r

(2nd term )  =    (3rd term ) / r

(2nd term)   =   (4/3)  / [  -2 / (3)^(2/3) ]  =  (4/3) * (3)^(2/3) / -2  = 

(4 / -2)  * ( 3^(2/3) / 3 )  =

-2 / (3)^1/3 

2x^2 -2x + 4 = 0       divide through by 2

x^2 - x + 2  = 0        subtract 2 from both sides

x^2 - x  =  - 2       complete the square on x

x^2 - x + 1/4  =  -2 + 1/4

(x - 1/2)^2  = -7/4       take both roots

x - 1/2  = ±i√7 / 2        add 1/2 to both sides

So

x = [ i√7 + 1] / 2     =  a

a^2  =  [ 1 + √7 i] [ 1 + √7 i ] / 4  =  [ 1 + 2√7 i - 7] / 4  =  [ √7 i - 3 ] / 2

Or

x = [ -i√7 + 1] / 2  =    b

b^2 =  [ 1 - √7 i ]  [ 1 - √7 i ] / 4 =  [ 1 -2√7 i  -  7 ] / 4  =  [ -√7 i - 3] / 2

a^2 + b^2   =   -6/ 2  =  -3

ab  =   [  1 +  i√7 ] / 2  *  [  1  - i√7 ] / 2   =  [ 1 + 7] / 4  =  2

So    ( x - a/b)  (x - b/a)  =

x^2 - (a/b)x   - (b/a)x +  (a/b)(b/a)  =

x^2 - ( a/b + b/a)x + 1   =

x^2 - ( a^2 + b^2) x / ab + 1  =

x^2 -  (-3)x / 2 + 1   =

x^2 + (3/2)x + 1

Let the side of the tetrahedron   =   s

Since  each face is an equilateral triangle......the slant height, h is given by

sin (60°)  = h / s ⇒     h = * sin (60°)  =  √3/2 * s = √3/2 

And he distance from M to the center of the triangle is given by

tan (30°) = C / [ (1/2)s ] ⇒  C = (1/2)s * (1/√3)  =  1 / [ 2√3]  =  √3/6 *s   (1)

And....using the Pythagorean Theorem, we can find the height of the triangle as

√ [  ( √3/2* s)^2  - (√3/6*s)^2 ]  =

s√ [ 3/4  - 3/36 ]  =

s√ [ 27 - 3] /36] =

s√[24/36] =

s√[2/3]     (2)

So....the tangent of AMB  =  (2) / (1)  =  s√[2/3]  / [ √3/6 *s] =

√2              6               6√2          

__    x      ___   =      _____  =       2√2     

√3             √3               3 

 Hi Melody, Thanks for your update....you are very always helpful...I can vouch for that!

I have a partial solution which I will post later on tomorrow (loading is a bit tough)...but need to rework  as in desperation I worked back from the answer, but  still not perfect.      

 Two consecutive numbers means one number is even, and one is odd.

 \(even\times even=even\) and \(odd\times odd=odd\) .

 Now, \(even-odd\) or \(odd-even\) both equal odd, so Tina is right.

You are very welcome!

:P

Can you clarify your question? Take the first one. What is 2,300, 7 , 9%. Do you want the present value of an annuity that pays $2,300 per year for 7 years @ 9% annual rate?

If f(x) = x2 + 5x, find limΔx-->0 [(f(x+Δx)-f(x)) / Δx]

Hi Old Timer,

We are no ignoring your question (part 2 that is).  Chris and I just have not worked out how to do it yet.  

Hopefully Heueka or Alan or Tiggsy or some other clever member or guest will show us all how it is done (hopefully in a nice easy to follow fashion.) 

Edit:

I hope I haven't offended anyone by using names. Maybe Rom or ElectricPavlov or Omi or geno or  ...   We have lots of very knowledgable people here now days.