ElectricPavlov

Just as an FYI :

    12.6 is three significant figures      3.8 is two significant figures

                               by the 'rules' the answer can have no more than two significant figures

                                 so if using significant figures, the answer would be 48 pounds

Sorry....  my mistake      all of the math is correct  EXCEPT h = +4      NOT - 4  

Thanx Helpbot for pointing that out to me !

l * w = area

1.2 l  *  1.12 w = 1.344  l*w      1.344 means the area increases by 34.4 %

Common ratio  r = 1/16

  converging series because   r <1

     sum =  a₁ /(1-r) = 2.1333...

Y= -2x^2+12x-8      Using x = -b/2a        Where b= 12    a= -2

7744     88²

y=(x-3+sqrt5)(x-3-sqrt5)   multiply this out :

y=x^2 -3x-sqrt5x - 3x +9+3sqrt5+ sqrt5x-3sqrt5 -5      simplify

y=x^2 -6x +4       multiply by -2

y = -2x^2+12x-8

x value of vertex =  - b/2a      use this value of x in the equation to find the y value of the vertex

2 (x-4)^2 +8      expand to find  a    b   c

2x^2 -16x + 40           a = 2    b = -16   c = 40

then    3a = 6    3b = -48    3c = 120

6x^2 -48x +  120

6 (x-4)^2  +24                 h = -4

Which makes sense, because all you did was multiply the original expression by 3

3 * (2(x-4)^2 + 8) =    6(x-4)^2+24     h = -4

If I understand correctly

1 element sets   = 7 c 1  =  7

2 element sets   = 7 c 2  = 21

3                                        35

4                                        35

5                                        21

6                                         7

7                                         1                Add them all together..........

Question: At an altitude of 15,000 meters above sea level in standard atmospheric conditions; a) What is the temperature value? (Temperature will be calculated in ℃.) B) What is the pressure value? c) What is the density value?

STP is sea level    T = 0^o C       or  273.15 K

standard lapse rate for temperature is approx  9.8 degree C per 1000 m

0 - 9.8(15) =  - 147^o C

   if you are using 15^o C for standard temp at sea level     the answer would be   -132 degrees C