Marine fuel is a combination of gasoline and motor oil. The standard gasoline and oil mixture is about 98% gasoline and about 2% motor oil. The "break in" mixture for a new engine is about 96% gasoline and about 4% motor oil. a. Write an equation in standard form that models the possible combinations of each kind of mixture you can prepare using 6 gallons of gasoline. b. If you prepare 4 gallons of the "break in" mixture, how much gasoline will you have for the standard mixture?

With 6 gallons of gasoline you could obtain 6.12 gallons of standard mixture or 6.25 gallons of break in mixture. In turn, if 4 gallons of break in mixture were prepared, there would be 2.16 gallons of gasoline left to prepare standard mixture.

Since the standard gasoline and oil mixture is about 98% gasoline and about 2% motor oil, and the "break in" mixture for a new engine is about 96% gasoline and about 4% motor oil, to determine the possible combinations of each kind of mixture you can prepare using 6 gallons of gasoline and, if you prepare 4 gallons of the "break in" mixture, how much gasoline will you have for the standard mixture, the following calculations should be performed:

A)

Therefore, with 6 gallons of gasoline you could obtain 6.12 gallons of standard mixture or 6.25 gallons of break in mixture.

B)

Therefore, if 4 gallons of break in mixture were prepared, there would be 2.16 gallons of gasoline left to prepare standard mixture.

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0.98g+ 0.02m= 7.1224 gallons standard mixture

0.96g+ 0.04m= 6.25 gallons break oil

3.92 gallons of gasoline  is required for  4 gallons the standard mixture

Step-by-step explanation:

If we have to use 6 gallons of gasoline this means that 98%= 6 gallons and 2% would be 6/98*2

Gasoline: Motor oil

6              :    x

98        : 2

Using the product rule

x= 6*2/98= 0.1224 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the standard oil the equation would be

0.98g+ 0.02m= 7.1224 gallons standard mixture

For break in oil

Gasoline: Motor oil

6              :    x

96        : 4

Using the product rule

x= 6*4/96= 0.25 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the break oil the equation would be

0.96g+ 0.04m= 6.25 gallons break oil

Part b. For  4 gallons of the "break in" mixture

96% of the 4 gallons of the "break in" mixture is gasoline =  3.84 gallons of gasoline

4% of 4 = 0.16 gallons of motor oil

Now the standard mixture 4 gallons would contain 98 % of gasoline and 2 % of motor oil

98% of 4 gallons= 3.92 gallons of gasoline

2% of  4 gallons= 0.08 gallons of motor oil.

answer:

step-by-step explanation:

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: