To know how many gumballs will fit in the machine, the
volume of the minature and the machine should be calculated. Volume of sphere:
V = (4/3) pi* r^3
Volume of the machine r = 6 in
V = (4/3) pi* 6^3 = 904.77 in3
Volume of the miniature gumballs
V = (4/3) pi* (1/3)^3 = 0.1551 in3
Divide the two volume to get the number of gumballs that can
fill the machine
904.77 in3/ 0.1551 in3 = 5832
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Answer:
either A or D
Step-by-step explanation:
Answer:
The data that we have is:
"Adrian's backyard pool contains 6.4 gallons of water. Adrian will begin filling his pool at a rate of 4.1 gallons per second."
Then we can write the amount of water in Adrian's pool as a linear function:
A(t) = 6.4gal + (4.1gal/s)*t
Where t is our variable and represents time in seconds.
We also know that:
"Dale's backyard pool contains 66.4 gallons of water. Dale will begin draining his pool at a rate of 0.9 gallons per second. "
We can also model this with a linear function:
D(t) = 66.4 gal + (0.9gal/s)*t
Both pools will have the same amount of water when:
D(t) = A(t)
So we can find the value of t:
6.4gal + (4.1gal/s)*t = 66.4 gal + (0.9gal/s)*t
(4.1gal/s)*t - (0.9gal/s)*t = 66.4gal - 6.4gal
(3.2gal/s)*t = 60gal
t = 60gal/(3.2gal/s) = 18.75s
In 18.75 seconds both pools will have the same amount of water.
Answer: 50
Step-by-step explanation:
A = lw (Area=length x width) 7x21= 147
The area is 147.