ElectricPavlov

5 (x^2 - 6x+5)

5 (x-1)(x-5) = 0

or   (5x-5)(x-5)= 0     

or  (x-1) (5x-25) = 0

Area of parallelogram = b x h =  10 x 4  m²

You mean like   # tiles = x + 4       assuming the squares and triangles are both tiles.......

Make sure your calculator is in 'radian' mode:

2= 15 cos(pi/30(x-30))

2/15 = cos (pi/30(x-30))

arcos 2/15 = 1.437   =  pi/30(x-30)        Now can you solve for 'x' ?

You didn't give a range over which to find x      arcos 2/15  also =  -1.437

    so   -1.437 = pi/30 (x-30)       Solve for  'x'

Using Pythagorean Theorem to

find the distance from  0,0      to     10,14

find the distance          0,0       to     13,9        which is lesser?  

According to WolframAlpha  it is 20

https://www.wolframalpha.com/input/?i=%28x%5E4+%2Bx%5E3+%2Bx%5E2%2Bx%2B1%29%5E4

The point will be    (x,0) because it is on the x-axis

distance from  -2, 0      is  (distance formula)    (x - - 2)^2 = d^2

distance from   0,4                                             (x-0)^2 + ( 0-4^2) = d^2

The distances are equal

   (x+2)^2  = x^2 + 16

x^2 + 4x + 4 = x^2 + 16

4x = 12

x = 3                           the point is    3,0

A    Use the average     for  1 hour    (14 +7)/2 = 11.5 in³/hr

B not sure what this means   but I believe     11.5 in^3/hr x 1 hr = 11.5 in³

C.   Not sure what 'net change in water' means

          but maybe this      f(x) - g(x)      find values for x = 0    then subtract the values of  x = 1

              at 0   f(x)  = 14.11    g(x) = 4        14.11 - 4 = 10.11

              at 1 (f(x) = .....         g(x) = ......                   =.........

10.11 - ....... =  net flow rate change from 10 to 11

1 gm/ml  x   10 ml   = 10 gm        (see how the 'ml' cancels out?)      This occurs at 4^o  C    as the density ( 1 gm/ml)  changes with temp

angle A = angle J      86

         B=K                 105

         D = M                44                

         C = L 

86 + 105 + 44 = 235^o     All of the angles (of a quadrilateral)  must sum to 360    360 - 235 = 125^o  = C = L