X²+15x+36<0
at first solve quadratic equation
D=b²-4ac= 225-4*1*36= 81
x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3
we can write x²+15x+36<0 as (x+12)(x+3)<0
(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
1 case) x+12<0, and x+3>0,
x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set
or second case) x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3) they are crossing , so (-12, -3) is a solution of this inequality
Answer:
The equations have the same mathematical form:
S = A + V0 t - 1/2 g t^2
They are both quadratic equations in one variable (t here)
This equation expresses the distance traveled by an object in time t in the vertical direction - it does not necessarily refer to a graph
In the other equation, y can be considered a vertical coordinate equation on a graph and x would be the horizontal coordinate
Note that in one equation 3 terms are necessary to describe the displacement of an object in one direction
In the other 2 terms in the x-direction describe the displacement in the other or y-direction
If the remaining 4 letters can be anything, you have 2*26^4 = 913,952 possible 5-letter ham radio call signs.
Add all the angles and subtract that number from 360. missing angle: 87
