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Very nice Chris I like your method better. :)

1.

Set f(x) = g(x)

2x + 1 = 2x^2 + 1    rearrange as

2x^2 - 2x = 0         factor

2x ( x - 1) = 0

Setting each factor to x and solving for x gives that  x = 0   and x = 1

x= 3 is incorrect

2.  The intersection points  of the graphed functions (if any) will be the solution point(s)

f(x) =  5 / (x + 7)  - 8

Note that x = -7   makes a denominator = 0

So....this is the vertical asymptote

The rational  expression  has a lower degree polynomial / higher degree polynomial

This means that we have a horizontal asymptote at y = 0

But the "-8 " shifts this down 8 units

So....the horzontal asymptote  is  y = -8

I think this is

3 / [ 5x^5 ]

The domain is  everything but x = 0  [it makes the denominator = 0 ]

So ...domain is  (-inf, 0 ) u (0, inf)

The range is a little trickier

We have a lower degree polynomial / higher degree polynomial

Whenever this happens..in the absence of any other shift up/down...we will have a horizontal asymptote of y = 0

So...the range is  (-inf, 0) U (0, inf)

Second answer

Here's a graph : https://www.desmos.com/calculator/jaehkdqcim

Thank you so much! Your explanation was very helpful!

oh makes sense, thanks

Here's one way to do this

Here's a pic :

DE/AD = CB/AB

DE/4.5 =13/9

DE = 13(4.5)/9  = 6.5

So ....the area of triangle ADE = (1/2)(DE*AD = (1/2)6.5 * 4.5 = 14.625 units^2

And we can find X  as the intersection of two lines

The first has the equation  y = (4.5 /6.5)x   ⇒ y = (9/13)x    (1)

D= (0, 4.5)  C = (13, 0)   so the slope of the line connecting these two points is

-4.5 /13  = -9/26

So....the line through these these two points is y = (-9/26)x + 4.5     (2)

Setting (1) = (2)  to find the x cooordinate of X we have

(9/13)x = (-9/26)x + 4.5

[(18/26) + (9/26) ] x = 4.5

(27/26)x = 9/2

x = (26/27)(9/2)  = 13/3 

And y = (9/13)(13/3) = 3

So.....the area of triangle DXE = (1/2) (4.5 - 3)(6.5) = (1/2) (1.5)(6.5) = 4.875 units^2

So the area of AEXD = area of triangle ADE + area of triangle DXE =  14.625 + 4.875 = 19.5 units^2

And the area of triangle BXC = (1/2)(13)(3) = 39/2 = 19.5 units^2

So  the ratio of the areas =  19.5 / 19.5  =   1    ..... exactly the same areas!!!

9/180 = 1/20   = .05

Two points on the graph help you calculate slope   =     rise/run = (5-1)/(-2-0)  = -2

Slope perpindicular = - (1/-2) = 1/2

SLope intercept form of the line   y = 1/2 x + b   sub in the point given to find  b

                                                      6 = 1/2 (7) + b        b = 2 1/2

y = 1/2 x + 2 1/2

Put   x = -3  into the function and we have that

(-3)^4 - 9(-3)^3 - 5(-3)^2  - 3(-3) + 4 =

81 + 243 - 45 + 9 + 4 =

292  = the remainder

For f^-1(-3)    we have that either

2 - x  = -3                        or                                      2x - x^2 = -3

x = 5                                                                          x^2 - 2x - 3 = 0

not defined  for 2 - x                                                  (x - 3) ( x + 1) = 0

                                                                                   x = 3   or x = -1

                                                                                  only the first value is defined for 2x - x^2

For f^-1(0)   we have that either

2 - x = 0                         or                                       2x - x^2 = 0

x = 2- 0                                                                      x^2 - 2x = 0

x = 2                                                                          x(x - 2) = 0

not defined for    2 - x                                                x = 0   or x = 2

                                                                                   only the second value is defined for 2x - x^2

For f^-1(3)  we have that either

2 - x = 3

x = 2 -3

x = -1

This is defined for 2 - x

So

f^-1(-3) + f^-1(0) + f^-1(3) =

3  + 2   - 1   =     

4

Thanks, Melody...

Here's another way without using Calculus

Positioning the circles as Melody did :

The intersection of two circles at the left side of the figure occurs at  A =    (sqrt(3) , 3)

Similarly, the intersection of the two circles at the bottom occurs at ( 3, sqrt (3) )

So  the distance between these two points is  AD =   sqrt [ 24 - 12 sqrt (3) ] cm

Therefore...using symmetry....we can construct  a square  AFGD of this side with an area of  [ 24 - 12sqrt(3)] cm^2     (1)

Looking at the segment AE... the slope of this segment = 3/sqrt (3) = sqrt (3)

Looking at the slope of DE....the slope of this segment is sqrt (3) / 3 =  1 / sqrt (3)

So   arctan (sqrt (3) ) - arctan (1/sqrt(3) ) =  60° - 30°   =   angle AED = 30°

So  the area of sector AED is  (1/2)(12)(pi/6)  =   pi  cm^2

And the area of triangle AED = (1/2)(12)sin(30)  = 3 cm^2

So....the area between the sector and the triangle is  [pi - 3] cm^2

And  using symmetry....4 of these areas = [4pi -12] cm^2      (2)

So....the shaded area is  (1) - (2) =

[ 24 - 12sqrt (3)] - [ 4pi - 12 ] cm^2  =   [ 36 - 12sqrt (3) - 4pi] cm^2  (exact)   ≈ 2.65 cm^2  (rounded)

He made a tiny rocket and launched it in class

translate graph to the left (by 3 units)

translates the graph up  ( by 6 units )

compress the graph closer to the x axis    (the "1/4"  does this )

Acoording to the Order of Operations...what we really have is this

(something) - (something x something) +  (something)

It's obvious that we want to subtract the smallest possible value of  (something x something) from the largest possible value of the first "something"

The largest possible value of the first something is   6/3  = 2

The smallest possible value of the middle multiplication is 4/5 * 2/3  = 8/15

Which leaves 2/2  as the final added "something"  = 1

So we have

6/3 - (8/15) + 1

2 + 1 - 8/15

3 - 8/15 

37 /15

Mr. BB, your equation would be correct if you leave off the triple-factorial (!!!).

a =bc / (b - c) is eqivalent to a =- bc /(c - b) !!!   

I know you like these.  It does seem to be an ideal symbol for you –especially when used in this form: (!!!Mr. BB). Read as Triple Deranged Mr. BB

Until next time Mr. BB, keep up the Derangements!!!.        

GA

Just put  ' 10 ' in to the equation for ' x ' 

.08 (10^3) + 10^2 + 10 + .26  =  190. 26   kw      with wind 10 m/s

See the solution here:  https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6

Deleted.....misread the Q !!   .

It ROTATES one time with each ORBIT around the Earth.   That is why we can never see the 'Dark Side of the Moon'.

   One ORBIT takes 27.32 days .

thank you 

Yeppers.

Thank you. My teacher only taught sig figs with a video, and it didn't cover scentific notation.

Thankksss yououuuuuu sooo muchhhhhh 

translate the graph to the right

compress the graph closer to the x axis

reflect the graph over the x axis