Achemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. the premium antifreeze solution contains 90% pure antifreeze. the company wants to obtain 180 gallons of a mixture that contains 70% pure antifreeze. how many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

40 gallons of water and 140 gallons premium antifreeze

Step-by-step explanation:

Let p represent the number of gallons of premium antifreeze in the mix. Then 180-p is the number gallons of water. The amount of pure antifreeze in the mix is then ...

0.90p + 0·(180 -p) = 0.70·180

p = 0.70·180/0.90 = 140 . . . . divide by the coefficient of p

180-p = 40 . . . . . find the water content

40 gallons of water and 140 gallons of premium antifreeze are in the mix.

step-by-step explanation:

the given growth  function for the length of the fish (cm) is

f(m) = 5(1.07)m = 5.35m

where 

m = months

at the end of the study, the length is 9.19 cm. the  time for conducting the study is determined as

5.35m = 9.19

m = 1.7178 months

part a.

the study was performed over a time period  of about 1.72 months. therefore  the domain for  plotting the function is 0 < = m < = 1.72.

note:

if growth  continues at this rate for up to m months, then the domain would be 0< =m< =m.

part b.

the growth function is a linear function.

when m=0, f=0.

therefore the initial growth is zero, which represents the y-intercept.

part c.

assume that the growth function is valid for 0< =m< =9.

the average rate of change of the function is the slope of the linear function, which is 5.35 cm/month.