Answer is:
x ≥ -18/5
2+4/9x ≥ 4+x
4/9x-x ≥4-2
4/9x-9/9x ≥2
- 5/9x ≥2 multiple by 9
-5x ≥2•9
-5x ≥18
x ≥ -5/18
Answer:
252 in²
Step-by-step explanation:
Recall area of regular polygon is A= (1/2)*(perimeter)*(apothem)
Area = (1/2)*(63)*(7)
= 252 in²
Answer:
r = 21
Step-by-step explanation:
nth of sequence P :

: 
Answer:
W = kq1q2 / r
Step-by-step explanation:
W varies jointly as the product of q1 and q2 and inversely as radius r
Product of q1 and q2 = q1q2
W = (k*q1"q2) / r
W = kq1q2 / r
Where,
W = work
q1 = particle 1
q2 = particle 2
r = radius
k = constant of proportionality
The answer is W = kq1q2 / r
Since f(x) is (strictly) increasing, we know that it is one-to-one and has an inverse f^(-1)(x). Then we can apply the inverse function theorem. Suppose f(a) = b and a = f^(-1)(b). By definition of inverse function, we have
f^(-1)(f(x)) = x
Differentiating with the chain rule gives
(f^(-1))'(f(x)) f'(x) = 1
so that
(f^(-1))'(f(x)) = 1/f'(x)
Let x = a; then
(f^(-1))'(f(a)) = 1/f'(a)
(f^(-1))'(b) = 1/f'(a)
In particular, we take a = 2 and b = 7; then
(f^(-1))'(7) = 1/f'(2) = 1/5