A woman who has recovered from a serious illness begins a diet regimen designed to get her back to a healthy weight. She currently weighs 103 pounds. She hopes each week to multiply her weight by 1.08 each week. (a) Find a formula for an exponential function that gives the woman's weight w, in pounds, after t weeks on the regimen. (b) How long will it be before she reaches her normal weight of 135 pounds?

a.) w = 103 * 1.08^t

b.) 3.5weeks

Step-by-step explanation:

If Her current weight is 103 pounds and she hopes to multiply her her weight each week by 1.08, then

her weight after 1 week = 103 * 1.08 = 103 * 1.08¹

Her weight after 2 weeks = [weight of week 1] * 1.08 = [103* 1.08] * 1.08 = 103 * 1.08²

Weight after 3 weeks= [weight of week 2] * 1.08 = [103 * 1.08 * 1.08] * 1.08 = 103 * 1.08³

Hence weight (W) after t weeks = 103 * 1.08^t

b.) If W = 135, Then

103 * 1.08^t = 135

1.08^t = 135/103

1.08^t = 1.31

Taking log of both sides,

log 1.08^t = log 1.31

t log 1.08 = log 1.32

t = log 1.32/log 1.08

t = 3.5 weeks.

Hence, it will take her 3½ weeks to get to 135pounds weight.

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2