Answer:
B.
Mn = Mn-1 + 125 for n > 1 ; M1 = 3,875
Step-by-step explanation:
Since they tell us after 1970, the first data would be that of 1971 and also that if we started since 1970, we do not know the data of 1969, therefore answer B is correct.
Replacing:
Let M1 = 1971 then Mn-1, that is M0 = 1970, we know that the population of 1970 the population is 3750, because it would be 125 less than the subsequent year, and in 1975 there are 3875. Therefore:
in n = 1
M1 = M0 + 125
3875 = 3750 + 125
this gives an equality, thus fulfilling the equation.
(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
I’m sorry I don’t see a diagram for this. Could you please scan one?
Answer:
Rounded to 73
Step-by-step explanation:
The exact number does not end so rounded to the closest whole number is 73
