Answer: 301.44
Explanation:
V = πr^2•h
3.14•4^2 • 6
Answer:
198.95 cm²
Step-by-step explanation:
Circumference = 50cm
C = πD
D = C/π
r = D/2
r = 50/π/2
r= 7.957747…..cm
Area = π r²
A = π x 7.957747….²
A = 198.95 cm² (2dp)
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Answer:
The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of
.
Step-by-step explanation:
The formula from the maximum distance of a projectile with initial height h=0, is:

Where
is the initial velocity.
In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is
. The critical points of the function are those who make
:


The critical value inside the interval is
.

The second step is to find the values of the function at the endpoints of the interval:

The biggest value of f is gived by
, therefore
is the absolute maximum.
In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of
.
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