Longer than forever.........!!!!!!!
THX, Melody !!!!!
Let 3^x = m .....so........(3^x)^2 = (3^2)^x = m^2 = (9)^x
So we have
m^2 - m + 1 complete the square on m
m^2 - m + 1/4 + 1 - 1/4
(m - 1/2)^2 + 3/4
And this will be a minimum of 3/4 when m = 1/2
Impressive, Melody !!!!!
y = x^2 + a y =ax
Set these equal
x^2 + a = ax
x^2 -ax + a = 0
The discriminant is
a^2 - 4a
To have real solutions.....this must be ≥ to 0....so
a^2 - 4a ≥ 0
a ( a - 4) ≥ 0
This will be true for a on these intervals : (-inf, 0 ] U [ 4, inf)
P(pen from the first box) = 4/8 =1/2
P ( crayon from the second box) = 7/12
So
P(pen from the first box and crayon from the second) = (1/2) (7/12) = 7/24
We have that
(36-20) take only Mathematics = 16
(27-20) take only Physics = 7
20 take both....
So (60) - ( 16 + 7 + 20) =
60 - 43 =
17 take neither....
Here : https://web2.0calc.com/questions/help_18684#r1
Sorry.....I meant "common ratio" instead of " common difference"
Ah, yes......I DO see that !!!!!
Thanks, Guest !!!!!
Also.....here.....
