The owner of a 4000-gal oil truck loads the truck with gasoline and keroseno. the profit on each gallon of gasoline is 16e and on each gallon of kerosene it is 12¢. how many gallons of each fuel did the owner load it
the profit was $620?
there
gallons of gasoline and
of kerosene.

There were 35 gallons of gasoline and 500 gallons of kerosene.

Step-by-step explanation:

Step 1: Write data for equation 1

Let gallons of kerosene be x

Let gallons of gasoline be y

Total capacity of truck is 4000

Step 2: Form equation 1

x + y = 4000

Step 3: Write data for equation 2

Profit of total gallons of kerosene is 0.12x

Profit of total gallons of gasoline is 16y

Total profit is $620

Step 4: Form equation 2

0.12x + 16y = 620

Step 5: Find x in terms of y from equation 1

x + y = 4000

x = 4000 - y

Step 6: Substitute value of x in equation 2

0.12x + 16y = 620

0.12 (4000 - y) + 16y = 620

480 - 12y + 16y = 620

4y = 620 - 480

y = 140/4

y = 35

Step 7: Substitute value of y in equation 1 to find x

0.12x + 16y = 620

0.12x + 16 ( 35 ) = 620

0.12x = 60

x = 500

There were 35 gallons of gasoline and 500 gallons of kerosene.

!!

1 3 5

step-by-step explanation:

i took the test: )

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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