a) 15cm^3
b) 8:16 am
1 liter = 1,000 cm cubic
Area A = 800
in 10 minutes
12 litters=12,000cm
12,000/800=15 cm
Tank A=15 cm ^3
Area B = 2,000
In 1 minute 5 liters = 5,000 cm^3
5,000/2000 = 2.5 cm^3 in 1 minute
To know how many minutes to reach 15:
15/2.5 = 6
6 minutes (from 8:10 am to 8:16 am)
5liters x 6 minutes =30 liters = 30,000 cm^3
30,000/2000 = 15 cm^3
Answer:
Write more compactly about functional relationships such as, “If you sell x tickets, then your profit will be ... Jason goes to an amusement park where he pays $8 admission and $2 per ride.
Answer:
Step-by-step explanation:
u=12kx^2
x^2=u/12k
...x=
(B) 8 less than the product of 3 and n
Here, we are given an equation 3n - 8
Let us consider each of the options one by one
(A) the product of 3, n and 8 will be written as 24n. Hence this isn't the correct answer.
(B) 8 less than the product of 3 and n will be written as 3n - 8. This matches our equation and hence is the correct answer.
(C) n minus 8 times 3 will be written as 24 - n. Hence, this is also incorrect.
(D) 3 times n less than 8 will be written as 8 - 3n. Thus, this is also eliminated.
Therefore, option B is the correct answer.
Learn more about expressing equations here-
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Answer:
1.) Zero ( 0 )
2.) 55.47 feet , 8.6 feet
3.) 17.2 feet
Step-by-step explanation:
The height, in feet, of the ball is given by the equation h(x)=−.25x2+4.3x, where x is the number of feet away from the golf club (along the ground) the ball is.
1.) Since the equation has no intercept,
The ball will start zero feet above the ground.
2.) The distance of the ball at the maximum height will be achieved by using the formula
X = -b/2a
Where b = 4.3, a = -0.25
Substitutes both into the formula
X = -4.3 / 2( - 0.25 )
X = - 4.3 / - 0.5
X = 8.6 feet
Substitute X into the function to get the maximum height
h(x) = −.25(8.6)^2 + 4.3(8.6)
h(x) = 18.49 + 36.98
h(x) = 55.47 feet
3) As the ball returns to the ground, the height will be equal to zero, therefore,
0 = -0.25x^2 + 4.3x
0.25x^2 = 4.3x
X = 4.3/0.25
X = 17.2 feet
The ball returns to the ground at about 17.2 feet away