You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:
<em>a(t)</em> = 40 ft/s²
<em>v(t)</em> = <em>v </em>(0) + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
<em>v(t)</em> = -20 ft/s + ∫₀ᵗ (40 ft/s²) d<em>u</em>
<em>v(t)</em> = -20 ft/s + (40 ft/s²) <em>t</em>
<em />
<em>s(t)</em> = <em>s </em>(0) + ∫₀ᵗ <em>v(u)</em> d<em>u</em>
<em>s(t)</em> = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) <em>u</em> ) d<em>u</em>
<em>s(t)</em> = 10 ft + (-20 ft/s) <em>t</em> + 1/2 (40 ft/s²) <em>t</em> ²
<em>s(t)</em> = 10 ft - (20 ft/s) <em>t</em> + (20 ft/s²) <em>t</em> ²
Answer:
Answer in file:
Step-by-step explanation:
Step-by-step explanation:
the Boscombe all very well football match well because I am 15 year
Answer:
Step-by-step explanation:


Volume = 
find partial derivatives using product rule

i.e.
Using maximum for partial derivatives, we equate first partial derivative to 0.
y=0 or x+y =6
x=0 or x+4y =12
Simplify to get y =2, x = 4
thus critical points are (4,2) (6,0) (0,3)
Of these D the II derivative test gives
D<0 only for (4,2)
Hence maximum volume is when x=4, y=2, z= 4/3
Max volume is = 4(2)(4/3) = 32/3
Answer:
1
Step-by-step explanation:
(−4)−(−2)–{(−5)–[(−7)+(−3)–(−8)]}
-4 + 2 - {-5 - [-7 - 3 + 8]}
-2 - [-5 + 7 + 3 - 8]
-2 - (-3)
-2 + 3
1