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 answer is 0.3 repeating or in a fraction 1/3
Step-by-step explanation:
All you have to do is divide 1/5 th by 6 and then make it a fraction it's not that hard.
It all depends upon what other things are given If hypotenuse is not given, in cosine or reciprocals are useless. In that case only tan and cot will be equally convenient if adjacent sides are given.
Answer:
160
Step-by-step explanation:
if girls is 6 and boys is 8
and there are 120 girls
120g= 6g * 20
meaning you multiply 8 by the same thing
160= 8b * 20
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