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You are correct, EP!!!!...I'll edit my answer  !!!

Suppose $f$ is a polynomial such that $f(0) = 47$, $f(1) = 32$, $f(2) = -13$, and $f(3)=16$. What is the sum of the coefficients of $f$?

Guess that we have a third degree polynomial of the form  ax^3 + bx^2 + cx + d

If  f(0)  =  47, then d =  47

And we have this system

a + b + c + 47  = 32  ⇒   a +  b + c   =  -15      (1)

8a + 4b + 2c + 47  = -13  ⇒  8a + 4b + 2c  = - 60   (2)

27a + 9b + 3c + 47  = 16  ⇒  27a + 9b + 3c  =  -31    (3)

Multiply (1)  by -2  add to (2)  we get

6a + 2b =  -30    ⇒  3a + b  = -15 ⇒  -9a - 3b = 45    (4)

Multiply (1)  by -3   add to (3)  we get

24a + 6b  = 14  ⇒  12a + 3b = 7   (5)

Add  (4)  and (5)  we get

3a =  52    ⇒  a    =  52/3

3(52/3) + b  = - 15

52 + b  =  -15

b =  -67

52/3   -  67 + c  = -15

52/3 + c  =  52

52 + 3c  = 156

3c  =  104

c  =  104/3

So

a + b + c + d  =

52/3  - 67  + 104/3  + 47

156/3  - 20  =

52  -  20   =  32

Hey CPhil....   What about the other math book?  5 history books + 1 math book = 6 books in the middle.

We have

Arc Length  =   2pi * r  * ( Given Angle  / 360 )     .....so...

24 pi  =   2 pi  * r *  (80 /360)        divide out pi

24  =  2 * r *  ( 2/9)

24  =  (4/9) * r      multiply both sides by  9/4

24 (9/4)  =  r   =  (24/4) * 9  =  6 * 9   =  54  in

We have

Arc Length  =  r * theta (in rads)      ....so......

12pi  =  r *  (4/7 pi)         divide out  "pi"

12  =  r  ( 4/7)         multiply both sides by  7/4

12 (7/4)  =  r   =   84/4  =  21 inches

What is the coefficient of x in (x^4 + x^3 + x^2 + x + 1)^4

Here's the expansion

x^16 + 4 x^15 + 10 x^14 + 20 x^13 + 35 x^12 + 52 x^11 + 68 x^10 + 80 x^9 + 85 x^8 + 80 x^7 + 68 x^6 + 52 x^5 + 35 x^4 + 20 x^3 + 10 x^2 + 4 x + 1

The  answer to the last question  is   20

Similar to how many ways can ten people line up while waitng for a bus:  10!

How many ways can 4 people (beads) line up on a bracelet?  4! = 24   BUT 1/2 of these will simply be 'reflections' and are disallowed.....so  24 x 1/2 = 12 ways.      (my guess....anyone else?)

There is a polynomial which, when multiplied by x^2 + 2x + 3, gives 2x^5 + 3x^4 + 8x^3 + 8x^2 + 18x + 9. What is that polynomial?

We can find this with polynomial division

                         2x^3   - x^2  + 4x  + 3

x^2 + 2x + 3  [  2x^5 + 3x^4 + 8x^3 + 8x^2 + 18x + 9  ]

                         2x^5 + 4x^4 + 6x^3

                         _____________________________

                                  - 1x^4  + 2x^3 + 8x^2

                                  - 1x^4  -  2x^3  - 3x^2

                                  __________________

                                                4x^3 + 11x^2 + 18x

                                                4x^3 +  8x^2  + 12x

                                                _________________

                                                            3x^2  + 6x  + 9

                                                            3x^2  + 6x  + 9

                                                            _____________

                                                                                 0

Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. What is the degree of $f(x) • g(x)$?

The degree will  be the result of the product of  x^4 * 2x^4  =  2x^8  ⇒   degree 8   

Find $t$ if the expansion of the product of $x^3 - 4x^2 + 2x - 5$ and $x^2 + tx - 7$ has no $x^2$ term.

The  x^2 term in the product  will be given by

( 28 + 2t - 5)      so   

28 + 2t - 5  = 0

2t  + 23  = 0

2t  = -23    ⇒    t   =  -23/2 

1)  In how many ways can I arrange 3 different math books and 5 different history books on my bookshelf, if I require there to be a math book on both ends?

We have   P(3,2)  =  6 ways to arrange the math books  and 6! ways of arranging the 1 math book and 5 history books in the middle....  so...6 x 6!  =   6 x 720  =  4320 ways

2)  I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue. In how many ways can I choose a pair of socks, provided that I get two socks of different colors?

We  can  either have   

White - Brown  =  4C1 x 4C1  =  4 x 4  =  16 ways

Brown - Blue  =  4C1 x 2C1   =   4 x 2   = 8  ways

White -Blue  = 4C1  x 2C1  =   4 x 2  =  8 ways

So...the total ways =   16 + 8 + 8  =   32

  ( EDIT )

Your keyboard seems to have a problem with the apostrophe key.

I think:

1.

There are 6 different ways a math book can be at both ends.

That leaves 6 books in the middle that can be arranged 6! ways

6 x 6! = 4320

Let $f(x) = x^3 + 3x ^2 + 4x - 7$ and $g(x) = -7x^4 + 5x^3 +x^2 - 7$. What is the coefficient of $x^3$ in the sum $f(x) + g(x)$?

The x^3 coefficient will have a sum of  1 + 5  = 6 

Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $a$ be a constant. What is the largest possible degree of $f(x) + a•g(x)$?

Depends upon the value of "a"

Note that if   a  = -1/2, the sum of the first two terms in each polynomial is 0....and the sum  of the second two terms in each polynomial is also 0....so the sum will produce a degree of 1.....any other value of "a" will produce a 4th degree polynomial

Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $b$ be a constant. What is the smallest possible degree of the polynomial $f(x) + b•g(x)$?

Just like the last, if "b"  is -1/2,  the smallest possible degree will be 1 

Suppose $f$ is a polynomial such that $f(0) = 47$, $f(1) = 32$, $f(2) = -13$, and $f(3)=16$. What is the sum of the coefficients of $f$?

Guess that we have a third degree polynomial of the form  ax^3 + bx^2 + cx + d

If  f(0)  =  47, then d =  47

And we have this system

a + b + c + 47  = 32

8a + 4b + 2c + 47  = -13

27a + 9b + 3c + 47  = 16

The solution to this is    a = 52/3, b = -67, c = 104/3, d = 47

And the sum of these is    32 

There are MANY sites on the internet that will give you the number of Primes between any two numbers up to a limit!. As it happens, I found that there are 8,363 Prime numbers between 10,000 and 100,000, which is a Prime number itself !! That was easy enough!.

We have 10 ways to choose each value  in the first row....so the total possibilities are

10 x 10 x 10 x 10 x 10   =  10^5   =   100,000

1.A telegraph has x arms and each arm is capable of (x-1) distinct positions , including the position of rest. The total no. of signals that can be made is?

My answer      x(x-1)

Your given answer:      

 (x-1 )^x  - 1

Ok I was assuming that you chose one for each but it seems you can choose many more than that.

It must be more like question 3.

I really do not understand the question properly. Maybe I will think about it more later when I have more time. Can you explain what the question is better Rosala?

Since you have excess Cl , the limiting factor is the H

16.2 gm of H2 is still 16.2 gm of H

Each 1.008 of hydrogen requires 35.45 gm of chlorine to form HCl

16.2 /1.008   x  35.45 = 569.732 gm Cl required

16.2 gm H + 569.732 gm of Cl = 585 .932 gm HCl

Volume of a sphere = 4/3 * pi * r^3       r = radius of sphere      pi~ 3.14

If you use the radius in inches, your result will be in cubic inches (in^3)

1.     cos(x)     now stretch it horizontally by 4

        cos (x/4)    Now shift it UP by '1'

         cos(x/4) +1 

2.   amplitude = 3/4

3. A cos (x ) wave with amplitude  1/2     has midline at 0       Now shift it UP by 5....the midline now shifts up by 5

      midline is at 5     SO:     y=5

1.  Minimum value is  y  = -6

2.  In the form    y  =  Asin(Bx) + C

A  =  5

Since the y intercept  is  (0, - 2), C  is  -2

To find B we need to solve this

B  =  [ 2pi ] / period  =  [ 2pi  ] / [ 6 pi]  =  1/3

So....the function is :

y = 5sin ( (1/3) x )  - 2

Here's the graph :  https://www.desmos.com/calculator/0s8smsawk2

The midlne is  y  = -2

The first point on the midline  is  (0, - 2)

The next point  [to the right of the first point ]  is a max at  (3pi/2, 3)

For the first question, the period is '2'    f= 1/period, so the frequency is  ' 1/2 '

For the second question, the period is pi     f= 1/period =   1/pi

First one  :

In the form    Acos(Bx) + C

We can  find this as follows

B  =  pi

B   =  2pi / period   

pi  = 2pi/ period      rearrange

period  =  2pi / pi    =  2

Andd the frequency  is     1  / period  =   1 / 2

Second one :

The period is   π

The frequency  is  :     1  / period    =   1  / π

(b)  P( x < 4)  =     ( 5 + 10 + 2 + 9 )  / ( 5 + 10 + 2 + 9 + 33 + 12 + 15 + 3 + 1 ) =  13/45

(c)  P(x  = 6)  =  15 / (5 + 10 + 2 + 9 + 33 + 12 + 15 + 3 + 1)  =  1/6

(d) P( x ≥ 5 ) =  (12+ 15 + 3 + 1) /  (5 + 10 + 2 + 9 + 33 + 12 + 15 + 3 + 1) =  7/30