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HAHAHA!!!....that's WAY easier than my clumsy approach....thanks, EP!!!

Can you please help me?.. Hope you had a great time off!

Determine a constant k such that the polynomial P(x,y,z)=x^5+y^5+z^5+k(x^3+y^3+z^3)(x^2+y^2+z^2) is divisible by x+y+z.

Uhhhh......  a slug is a unit of MASS  , not average velocity

  so   slug/in^3   would be a measure of density....

Not sure about this one Julius.....thinking like you, that there is something there we aren't seeing.....seems too simple, and something is going to jump up and bite us....but I would say

Lim as x--->0  of this is equal to  (1-0)/(1+0) = 1....     Hope it is correct!

No it’s the same in physics and math.  You’re confused because in the world of mor.onic stup.idity  it’s slugs per in³.

Thank you, I do get it. I appreciate your help. :)

noppppppppppppppppppppppppppppppppppeeeeeeeeeeeeeee

As before (do you 'get' this?)

(x-1)   moves the graph 1 unit to the RIGHT

+2       moves the graph UP 2 units 

     ANSWER:  A

First one: slightly different approach...see diagram   the 1's and 4's were given in the question for the squares,

1)

P(x)   =   55,000√[x - 1945]

In which year was the population 275,000  ?

What is the value of  x  when  P(x)  =  275,000   ?

275,000   =   55,000√[x - 1945]

                                                     Divide both sides by  55,000 .

5   =   √[x - 1945]

                                             Square both sides of the equation.

25   =   x - 1945

                                             Add  1945  to both sides.

1970   =   x

In the year  1970  , the population was  275,000 .

2)

\(f(x)\,=\,\sqrt[3]{x-2}\)

                                       Instead of  f(x)  let's write  y .

\(y\,=\,\sqrt[3]{x-2}\)

                                       Now let's solve this equation for  x . Raise both sides to the power of  3

\(y^3\,=\,x-2\)

                                       Add  2  to both sides.

\(y^3+2\,=\,x \\~\\ x\,=\,y^3+2\)

                                       So the inverse function is...

\(f^{-1}(x)\,=\,x^3+2\)

Yep......looking at the graph is a neccessity here....LOL!!!!

Oh whoops haha I actually thought it was C not D, and no wonder I didn't get the answer I didn't even really look at the graph. Thanks~

Velocity  and average velocity are in the same units

Could be   m/s   or   km/s   or m/hr  or ft/s   etc     (distance units) / (time units)

Definitely not D  !!!....the graph shows it goes higher than this

Each little  "tick"  mark on "y"  represents 4 ft

If we go to 1.5 seconds on x.....note that the that the height is about 1 tick mark past 96 ft at this time

So   ...  it reaches a height of 96 + 4  =   100 ft   at 1.5 sec

Looks like   "a"  might be the best option   !!!

Do you see this, RP??? 

2. We can  find AB  thusly

sin (60)  = AB / 24

24 * √3 /2  =  AB

12√3  = AB

And  BE  =  (1/2)  of   AE  = 12

But ABE  and  BCE form similar triangles

So....if  BE  is  (1/2) of AE

Then BC  is  (1/2)  of AB  = 6√3

And CE  is  1/2 of BE  =  6

And triangle DCE is similar to triangle CBE

So since CE  is (1/2) of BE....then...

ED  is(1/2)  of CE  = 3     and

DC  is (1/2) BC  = 3√3

So  the perimeter of   ABCD  =

AB + BC + CD  + DE + EA  =

12√3  + 6√3  + 3√3  + 3  + 24  =

 [ 21√3  +   27 ] cm  ≈  63.4 cm

If I haven't made any errors  !!!!

Solve for x:

sqrt(120 - sqrt(x)) = 0

Square both sides:

120 - sqrt(x) = 0

Subtract 120 from both sides:

-sqrt(x) = -120

Multiply both sides by -1:

sqrt(x) = 120

Raise both sides to the power of two:

x = 14,400

1.  

The side of the smaller square  = 1 cm   and the side of the larger square  = 4 cm

Let the distance from  the top right of the smaller square to the intersection of AB with the side of the larger square  = x

So  the small triangle formed above the smaller square has a hypotenuse of   √[ 1 + x^2]

And  the larger triangle formed at the "top" part  of the larger square has  legs of  4  and [ (4) - (1 + x)]  =  (3 -x)    so    the hypotenuse  of this triangle  is  √  [ 4^2  + (3 - x)^2 ]  =

√ [  16  + 9 - 6x  + x^2 ]  =  √ [ x^2 - 6x + 25 ]

And the triangles are similar  such that

1 / √[ 1 + x^2]  =  4 /  √ [ x^2 - 6x + 25 ]      cross-multiply

√ [ x^2 - 6x + 25 ]  =  4 √[ 1 + x^2]          square both sides

x^2  - 6x + 25   =  16 (x^2 + 1)

x^2  - 6x + 25  =  16x^2  + 16

15x^2 + 6x - 9  =  0

5x^2  + 2x - 3  =  0

(5x - 3) (x + 1)  =  0

Setting each factor to 0 and solving for x produces   x  = 3/5  = .6  and x  = -1

But  x  is a length so it can't be negative

So  x  =  3/5 cm = .6cm

So....the length of AB  is

√[ 1 + (.6)^2]   +   √ [ (.6)^2 - 6(.6) + 25 ]  =

√ [ 1.36]  +  √ [ 21.76 ]  ≈ 5.8  cm    

Start in the middle and work your way outward

|-1+1| = 0

|-0-1| =  |-1|=1

|1+1|=2

x+3   shifts it  3 to the LEFT

-3  stretches it by -3 and FLIPS (reflects) it around the x- axis

+5 shifts it UP 5 units

Answer : A

A would be the answer

[6C2 * 6] / 6^3 = 90/216 =5 / 12

Tysm ^-^

We don't need any math for the first one, NSS

We can never get a negative answer out of a positive root

Thus....all the answers that have negative x values included are no good

So....by process of elimination....the last answer is correct  !!!

Second one

We have this

0.75 ( 140)  = s^2  /  [ 30.25 ]      multiply both sides by  30.25

(30.25) ( 0.75) (140)  =  s^2

Take  the positive root of both sides

√ [ (30.25) ( 0.75) (140)  ]  =  s  ≈   56 mph 

First find the vertex.....can you tell me what that is?

From the graph the vertex is   h,k (or the x,y point) =  -3,4

Vertex form of parabola......we answered that in your las question....remember it?

y = a(x-h)^2 +k   is vertex form      Now let's put  -3 , 4 in there

y= a (x-(-3))^2 + 4       Now how can we find the value of 'a' ?  find a convenient point on the graph (not the vertex)     ....look at x = -5       at that point y = -4    Sub this in to the working equation then solve for 'a'

-4 = a (-5 +3)^2 +4        means 'a' = -2   soooooo the equation becomes

y = -2(x+3)^2 +4

Thank you so much!!