Help! (I will give you brainliest!!!!)

If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n

natali 33 [55]

Answer:

$f^{-1}=3ln_2(\frac{t}{100})$

f^-1 represents the number of hours the bacteria developed

Step-by-step explanation:

$f(t) = 100 \cdot 2^\frac{t}{3}$

To find inverse of a function , replace f(t) with y

$y= 100 \cdot 2^\frac{t}{3}$

replace t with y and y with x

$x= 100 \cdot 2^\frac{y}{3}$

solve for y

divide both sides by 100

$\frac{x}{100} =2^\frac{y}{3}$

take ln on both sides

$ln \frac{x}{100} =\frac{y}{3}ln 2$

divide both sides by ln

$ln_2(\frac{x}{100}) =\frac{y}{3}$

multiply by 3 on both sides

$y=3ln_2(\frac{x}{100})$

Replace x with t and y with f^-1

$f^{-1}=3ln_2(\frac{t}{100})$

f^-1 represents the number of hours the bacteria developed

7 0

2 years ago

How do i find the sum or difference of this fraction

gregori [183]

1st use the Step! Please, Excuse, My, Dear, Aunt, Sally! Ehich also means (PEMDAS) Parenthesis (), Exponents ", Multiple ×, Division÷, Addition+, and Subtraction-. Than your going to solve! (3n+5)×9=n×7
27n+45=7n
27n-7n=-45
20n=-45
n=-45/20= -9/4

7 0

2 years ago

You are given g(x)=4x^2 + 2x and

Strike441 [17]

Answer:

324

Step-by-step explanation:

Given:

$g(x)=4x^2+2x\\ \\f(x)=\int\limits^x_0 {g(t)} \, dt$

Find:

$f(6)$

First, find f(x):

$f(x)\\ \\=\int\limits^x_0 {g(t)} \, dt\\ \\=\int\limits^x_0 {(4t^2+2t)} \, dt\\ \\=\left(4\cdot \dfrac{t^3}{3}+2\cdot \dfrac{t^2}{2}\right)\big|\limits^x_0\\ \\=\left(\dfrac{4t^3}{3}+t^2\right)\big|\limits^x_0\\ \\= \left(\dfrac{4x^3}{3}+x^2\right)-\left(\dfrac{4\cdot 0^3}{3}+0^2\right)\\ \\=\dfrac{4x^3}{3}+x^2$