Answer:
20 gallons were used from the tank with the octane rating of 80.
Step-by-step explanation:
Given : One tank of gasoline has an octane rating of 140 and another tank of gasoline has an octane rating of 80. To obtain a mixture of 60 gallons with an octane rating of 120.
To find : How many gallons should be used from the tank with the octane rating of 80?
Solution : Let x be the volume of gas whose octane rating is 140 and
y be the volume of gas whose octane rating is 80.
Then, we form the equation as:
(1) 
or 
.........[3]
(2) 
Now multiply equation (2) by 4
...........[4]
Subtract (4) from (3)
Put x in equation [4] we get,
Therefore, x=40 and y=20
So, 20 gallons were used from the tank with the octane rating of 80.
Answer:
y=9
Step-by-step explanation:
∠CAE = 120°
∠CAD = 60°
∠BAE = 180°
∠DEC = 30°
We start out with the fact that points C and D split the semicircle into 3 sections. This means that ∠BAC, ∠CAD and ∠DAE are all 60° (180/3 = 60).
Since it forms a straight line, ∠BAE is 180°.
Since it is formed by ∠CAD and ∠DAE, ∠CAE = 60+60 = 120°.
We know that an inscribed angle is 1/2 of the corresponding arc; since CD is 1/3 of the circle, it is 1/3(180) = 60; and this means that ∠DEC = 30°.
1110.5 is the answer and your welcome
Answer:
<em>96π units²</em>
Step-by-step explanation:
Find the diagram attached
Area of a sector is expressed as;
Area of a sector = θ/2π * πr²
Given
θ = 3π/4
r = 16
Substitute into the formula
area of the sector = (3π/4)/2π * π(16)²
area of the sector = 3π/8π * 256π
area of the sector = 3/8 * 256π
area of the sector = 3 * 32π
<em>area of the sector =96π units²</em>
