We need to convert the mileage from mi/gal units into to km/L units using the conversion factors.
(31.0 mi/gal) x (1 km / 0.6214 mi) x (1 gal / 3.78 L) = 13.20 km/L
Next, we divide the distance by the mileage.
(142 km) / (13.20 km/L) = 10.79 L
<span>Therefore, you need 10.79 liters of gasoline.</span>
L = 2 W
B = L x W = 2 W²
Side Area = 2 W H + 2 L H = 2 H ( W + L ) = 6 H W
V = 2 W² H = 10
H = 5 / W²
Cost = 15 * 2 W² + 9 * 5/W
= 30 W² + 270/ W
C ` = 60 W - 270 / W²
= ( 60 W² - 270 ) / W² = 0
60 W² = 270
W ² = 270 : 60
W² = 4.5
W = √ 4.5 = 2.12
Cost (min) = 15 * 2 * 4.5 + 30 / 2.12 = 135 + 14.15 = $149.15
Answer: The cost of materials for the cheapest such container is $149.15.
X - 9 < 0
add 9 to both sides:-
x - 9 + 9 < 9
x < 9
x - 9 is negative for all values of x less that 9.
Answer:
step 2
Step-by-step explanation:
The student first made an error in the step 2.
Because if we simplify the equation A in step 1, we should distribute the right side of the equation following the rule a(b-c) = a*b +a*(-c)
So, it would be
–7y = (-7)*(16)+(-7)*(-2z)
=> -7y = -112 + 14z
9514 1404 393
Answer:
60
Step-by-step explanation:
Let m represent the mark Cheryn scored on her first quiz. Then her average is ...
(m +76 +89)/3 ≥ 75
m +165 ≥ 225 . . . . . . . . multiply by 3
m ≥ 60 . . . . . . . . . . . . subtract 165
Cheryn scored a minimum of 60 marks on her first quiz.
