CPhill

Here's my best effort....

Since a checker is placed on a diagonal....we have 8  possibilities  for the checker's placement

If we place the checker on Row 1, Column 1   we have  64 total checkers  = 8^2

If we pllace the checker on Row2, Column 2, we have 49 total checkers  = 7^2

Following this pattern, the expected value is

[8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2 ] / 8  =   204 / 8 =  51/2 =  25.5 checkers

Note that  in each  "piecewise"  part of the function, the  greatest value of  [f(x)  - x]  occurs at the left part of the interval  and is   0

And note that the least value of  f(x)  - x  occurs at the right part of the piecewise interval and is  "almost"  -1

So...the range of  f(x)  - x   is  ( -1, 0 ] 

I'm not sure about this, but here's my take

Note that any two elements drawn from the  set   { 60, 61, 62, 63,......,98, 99, 100}  will always have a  sum > 120  

And  the set  {1, 2, 3,.....,17, 18, 19}   can be combined with the first set such that no two elements drawn from this "combined" set will have a sum of 120

So...the sum of the elements of this set will be :

(19)(20) / 2    +   [100 + 61] * 40/2 + 60

190  + 161* 20 + 60   =

190 + 3220 + 60 =

3470

3.Find the equations for two lines that both go through the point (1,2), one with slope 3 and one with slope -1/3. (write the numerical coefficient of each term to complete the required equation) 
the answer would be.
a. With the slope =3: __ x-y- ____=0  (my answer y=3x-1)
b. With slope = -1/3: x + ___ y - ___=0 (my answer y=-1/3x+7/3)

First line

y = (3)(x - 1) + 2

y = 3x - 3 + 2

y = 3x  -  1      subtract y from both sides and rearrange as

3x - y - 1  = 0   [ your answewr is correct in slope-intercept form ]

Second line

y= (-1/3) ( x - 1) + 2     multiply through by 3

3y = (-1)(x - 1) + 6

3y  = -x + 1 + 6

3y  = -x + 7       add x to both sides, subtract 7 from both sides

x + 3y  - 7   = 0

2.Find the equation of the lines that are both passing through the point (10,-8) with slopes 5 and -1/5. 
Answer: ___ x - y - ____ = 0  (  my answer y=5x-58)
x + ___ y+ ____ = 0   (my answer y=-y=-1/5x-6)

First line

y = 5(x - 10) +  -8

y = 5x - 50 - 8

y = 5x - 58       subtract y from both sides

5x - y - 58  = 0     [  this is the same as your answer, just in "standard form" ]

Second line

y  = (-1/5) ( x - 10) + -8       multiply through by 5

5y  = (-1)(x - 10)  +  - 40  simplify

5y  = -x + 10  - 40

5y  = -x - 30      add x, 30 to both sides

x + 5y  + 30  = 0  [ your answer is in slope-intercept form and is correct.....I think they want "standard form" ]

1.Find the equation of the line which contains the point (-2,3) and is parallel to the line 3x+5y=17. (write the numerical coefficient of each term to complete the required equation) 
Answer would be: ___x + ___ y - ___ = 0 

Let us find the slope of the given line...in the form Ax + By  = C, the slope will be given by   -A/B

So...the slope of the given line  =  -3/5

And since parallel lines have the same slope, we can  write an equation of a line having a slope of -3/5 and passing through ( -2,3) thusly

y  =(-3/5)(x -  -2)  + 3      simplify

y = (-3/5) ( x + 2)  + 3      multiply through by 5

5y  = -3(x + 2) + 15

5y = -3x - 6 + 15

5y = -3x + 9     add 3x to  both sides, subtract 9 from both sides

3x + 5y - 9 = 0

Well...it  COULD be a right triangle with legs of 6950  and 3760  [ the hypotenuse will not be an integer value]

If so....the area  =    (1/2) (product of the legs)   = (1/2)(6950)(3760)  = 13726250 cm^2

Let us assume that  we can factor  x^4 + k  in the following manner :

(x^2  + ax  + b) (x^2 - ax  + b)

Expanding this, we have 

x^4 + ax^3 + bx^2

      -  ax^3 -a^2x^2 - abx

                  +bx^2   + abx   + b^2

___________________________

x^4      +   (2b - a^2)x^2         + b^2

Since the  x^2 term will "disappear", this implies that  b^2  = k

Then  b can possibly range from 1 - 44   since 1^2  = 1  and 44^2  = 1936.....note that 45^2  = 2025 which is too large

And since the  x^2 term disappears, it must be that  2b - a^2  =  0

Add a^2 to both sides  and we have that   2b  = a^2

Taking the square root of this we have that

√[2b]  = a

Since a is an integer, then  √[2b]  must also  be an integer

And the values that make √[2b]  an integer is when  b  = 2    b  = 8    b = 18   and b  = 32

So  (a, b)   =  ( 2,2)  (4, 8)   ( 6, 18)    and  ( 8, 32)

So........the following  product trinomials are possible

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

(x^2 + 4x + 8) (x^2 - 4x + 8) = x^2  + 64

(x^2 + 6x + 18) (x^2 - 6x + 18)   x^2 + 324

(x^2 + 8x  + 32) (x^2 - 8x + 32)  = x^2 + 1024

Yep...you are correct...I forgot that we  need to  find the integer values that make the inequality untrue...-4 should be included!!!

We can solve this by inspection

Note that when  x = 1   the expression  under the entire radical  is 0

And x cannot be smaller than this   (note if it were, for example, 0,   1 - √2  under the radical would be  negative

And  x  cannot be  larger than 4   because    [2 - √x ]   would result in a negative quantity under a radical

So...the domain  is  [ 1, 4 ]

Here's a graph : https://www.desmos.com/calculator/xxzra4wyai