ElectricPavlov

The complex plane looks like the usual cartesian coordinate plane....but the   x axis is for the REAL part of your complex number and the imaginary part is plotted on the ' i ' axis  (y-axis)....your graphed point would look like this (note the labels on the axes):

"The highest point in a particular section of a graph"

Here is a graph of a function with a lot of relative maximums:

Your first Q is missing some vital info.

2.  This is a bowl shaped parabola .....the minimum value for s   is      - b / 2a   = - (-8)/(2(1)) = 4   

                then subbing s = 4 into the eq   yields   2(4)^2-8(4)+19 = 19        4,19

3. This is a dome shaped parabola    max  is   -b/2a  = - (8)/(2(-1)) = 4 

                then subbing t=4 into the eq yields     -(4)^2 + 8t -4  = 12               4,12

7 / sec 50    =    7 / (1/cos 50) = 7 cos 50  = 7*cos(50) = 6.754762199444

Ideal gas law  is  PV =nRT      since n is constant (# moles) for this experiment, we can say PV = kT

and then   P/k = T/V       Pressure in the balloon is constant for all three balloons    so we can say that for all three balloons  P/k is constant  

So initially

k = T/V    =

V₁/T_{1 = 600/285 =2.11=k    for all of the balloons}

_{Now for the second balloon ' k ' has to be the same}

_{630/T2 =2.11   T2 = 298.6 Kelvin}

_{For balloon 3}

_{k = 2.11    = V3/T3 = V3/265    yields   V3 = 559 ml}    

Let's look at the vertex form of a parabola

th vertex   h, k  of this parabola is   (-3,0)

a (x-h)^2 +k  =  y         sub in h and k

a (x-(-3))^2 + 0 = y      now pick an easy point on the parabola.......like   -2,1 and sub it in to find 'a'

a (-2 +3)^2 = 1

a = 1

so:   (x+3)^2 = y    or   x^2 + 6x + 9 = y

in 3x+2     put in  x^2-5x-1   for 'x'  and see what you find.....

f(x) = 2/sqrt(x)      domain is all x that do not result in a denominator = 0

(-inf,0) U (0, + inf) 

f(x - g(x) + h(x)   =   2x^2+3x-9     - 5x-11    +-3x^2+1        gather like terms and simplify

                             - x^2 - 2x -19

No.....the summation calculators can only go so far....so I put in 99999 to 'imitate' infinity....then the calc spits out all of the calculations....

   the result was .9999900000000131      so I think it is safe to THINK/assume   ' 1 ' would be the answer.....