Let h=height of water
Let r=radius of water surface
r/h=16/8 =2, so r=2h.
The volume of water is:
v=(1/3)×π×r²×h
=(1/3)×π×(2h)²×h
=4/3πh³
dv/dh=4πh^2
By chain rule:
dv/dt=dv/dh×dh/dt
but
dv/dt=4
thus:
4=(4πh)×dh/dt
dh/dt=4/(4πh²)
when h=6cm we have:
dh/dt=4/(4π6²)
=0.00884 cm³/min
Step-by-step explanation:
o2+a2=h2
100} h^{2}=(10ft)^{2} +(17ft)^{2}
h2=100ft2+289ft2
h=389ft2−−−−−√
→19.7ft
this is an example on how to do it
8 a) y=x² stretched vertically by 8
That means (0,0) stays (0,0), and (1,1) becomes (1,8) and (-1,1) becomes (-1,8)
Answer: y = 8x²
8 b) y=x² compressed vertically by 1/5
Answer: y = (1/5) x²
9.
y = ax² + k through (-1,3) and (3,-13)
That means
3 = a (-1)² + k
-13 = a (3²) + k
Subtracting,
16 = a - 9a = -8a
a = -2
3 = -2 + k
k = 5
y = -2x² + 5
Check:
-2(-1)^2 + 5 = 3, good
-2(3)^2+5 = -18 + 5 = -13, good
Answer: a = -2, k = 5
Answer:
a) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
it would take 6 min to type 144 words
Step-by-step explanation:
72/144 = 3/6
