What is the b(t)=((t)1200)/(1+39e^{-0)0.9t} ?

The b(t)=((t)1200)/(1+39e^{-0)0.9t} is

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b(t)=((t)1200)/(1+39e^{-0)0.9t}

b (t) = \frac{( t ) 1200}{1 + 39 e ^{- 0} 0.9 t}

Domain : t < - \frac{10}{351} or t > - \frac{10}{351}

Range : f (t) < \frac{4000}{117} or f (t) > \frac{4000}{117}

X Intercepts : (0, 0), Y Intercepts : (0, 0)

Asymptotes : Vertical t = - \frac{10}{351}, Horizontal y = 34.18803 \dots

Extreme Points : None

Interval Notation

Domain : (- \infty, - \frac{10}{351}) \cup (- \frac{10}{351}, \infty)

Range : (- \infty, \frac{4000}{117}) \cup (\frac{4000}{117}, \infty)

Solution steps

Domain of

\frac{t 1200}{1 + 39 e ^{- 0} 0.9 t} : t < - \frac{10}{351} or t > - \frac{10}{351}

Range of

\frac{t 1200}{1 + 39 e ^{- 0} 0.9 t} : f (t) < \frac{4000}{117} or f (t) > \frac{4000}{117}

Axis interception points of

\frac{t 1200}{1 + 39 e ^{- 0} 0.9 t} :

X Intercepts

: (0, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{t 1200}{1 + 39 e ^{- 0} 0.9 t} :

Vertical

: t = - \frac{10}{351},

Horizontal

: y = 34.18803 \dots

Extreme Points of

\frac{t 1200}{1 + 39 e ^{- 0} 0.9 t} :

None

p (x) = 6 x^{4} + 5 x^{3} + 3 x^{2} - 3 x + 2

f (m) = m^{2} + 21 m + 109

y = - 0.3 x^{2} + 18 x - 100

x = 2 t^{3} + 4 t^{2} + t

f (x) = e^{3 x} + 64

What is the b(t)=((t)1200)/(1+39e^{-0)0.9t} ?
The b(t)=((t)1200)/(1+39e^{-0)0.9t} is