Im not sure but it should be 8.5
(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
Answer:
x=2
......
Step-by-step explanation:
x=3+1/x
x^2=4
x=2
Answer:
key is 18
exit tag is 15
lock is 14
key*tag*lock = 3780
Step-by-step explanation:
54/3 is 18 so each key is 18
48-key = 30 so each tag is 15 since 15+15 = 30
18*15*14=3780
Answer:
8.5 litres to millimeters
litres to millimeters
1 : 1000
8.5 litres = 8500 millimeters
