Answer:
COMPLETE QUESTION:
To the right are drawings of a wide and a narrow cylinder. The cylinders have equally spaced marks on them. Water is poured into the wide cylinder up to the 4th mark (see A). This water rises to the 6th mark when poured into the narrow cylinder (see B). Both cylinders are emptied, and water is poured into the narrow cylinder up to the 11th mark. How high would this water rise if it were poured into the empty wide cylinder?
a)To the 7 1/2 mark b)To the 9th mark c)To the 8th mark d)To the 7 1/3 mark
e)To the 11th mark
ANSWER : Option D (To the 7 1/3 mark)
Step-by-step explanation:
First part of the question enables us to get the relationship between the radius of the wider cylinder (R) and the narrow cylinder(r) i.e
Volume of cylinders
π x R² x 4 = πxr²x 6
R²/r² = 6/4
after both cylinder were emptied
π x R² x h = π x r² x 11
R²/r² = 6/4 = 11/h
h = (4 x 11) /6 = 22/3 = 7 1/3 mark
Therefore, the height of the water in the wide cylinder is 7 1/3
(2200 rev/min) x (360 deg/rev) x ( min/60 sec) = 13,200 degrees per second
(2200 rev/min) x (2 pi radians/rev) x (min / 60 sec) = <u>73 and 1/3 pi</u> radians/sec
Answer:
13, 14, 15
Step-by-step explanation:
Let x = age of one of the siblings
Since the age of the siblings are consecutive integers, the age of the next sibling would be x + 1 and the age of the third sibling would be x + 2
x +( x +1) + (x +2) = 42
3x + 3 = 42
collect like terms and solve for x
x = 13
The age of the first sibling is 13
The age of the second sibling is x + 1 = 13 + 1 = 14
The age of the third sibling is x + 2 = 13 + 2 = 15
Answer:
.
Step-by-step explanation:
