A patient requires an IV of D5W, a 5% solution of dextrose (sugar) in water. To the nearest milliliter, how much D20W, a 20% solution of dextrose in water, must be added to 100 milliliters of distilled water to produce a D5W solution? Please show all work.

Step-by-step explanation:

From the question we are told that

5% solution of dextrose sugar in water

20% solution of dextrose in

100 milliliters of distilled water water

Generally in mathematically solving we call A the amount of D20W added

Since 20% of dextrose is needed

Total quantity after addition (100+x)

for DW5% total solution is

Mathematically

Therefore the required quantity

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

step-by-step explanation:

let's assume

number of candles set =x

price of of x candles =p(x)

we are given

a candle maker prices one set of scented candles at $10 and sells an average of 200 sets each week

so, we get point as (200,10)

he finds that when he reduces the price by $1, he then sells 50 more candle sets each week

so, we can find slope

now, we can use point slope form of line

now, we can plug values

we can use point

and find b

now, we can plug it back

and we get

now, we can find revenue

we know that

revenue = price*quantity

so, we get