Answer:
Step-by-step explanation:
I assume that you mean
sec(x)-tan(x) = 1 / ( sec(x) + tan(x) ) , right ?
then this is equivalent to
[ sec(x) - tan(x) ] x [ sec(x) + tan(x) ] = 1
let s evaluate it, we got
sec2(x) - sec(x)tan(x) + - sec(x)tan(x) - tan2(x) = sec2(x) - tan2(x)
= (1 - sin2(x) ) / cos2(x) = cos2(x) / cos2(x) = 1
as cos2(x) + sin2(x) = 1
Basically we need to add and subtract 1/2 from 1/6:
(1) 1/6 - 1/2 = 1/6 - 3/6
= -2/6
= -1/3
(2) 1/6 + 1/2 = 1/6 + 3/6
= 4/6
= 2/3
Therefor the two numbers that are located 1/2 unit from 1/6 are -1/3 and 2/3
the answer issss −
1.9
b
+
7.8
I’m not sure but please tell me where i can watch it
For this case we have an equation of the form:
y = A * (b) ^ t
Where,
A: initial amount
b: decrease rate
t: time in years.
Substituting values we have:
7 = A * (0.5) ^ ((1/20) * t)
For t = 45 years we have:
7 = A * (0.5) ^ ((1/20) * 45)
Clearing A:
A = 7 / ((0.5) ^ ((1/20) * 45))
A = 33.3
Answer:
the density of the substance when it was deposited 45 years ago was:
A = 33.3 milligrams per square centimeter
