Answer:
all work is shown and pictured
Answer:
1. <em>d</em> - 6 ft
where <em>d</em> is the width
2. <em>t</em> + 6 h/week
where <em>t</em> is the amount of time per week Theodore studies
3. <em>t’</em> - 6 a
where <em>t’</em> is Tracy’s age
4. (<em>k</em>/3) - 2
where <em>k</em> is the points that the panthers scored
Step-by-step explanation:
Write the variable and do the operations as described!
“less than” means subtraction, “more […] than” means addition and “one-third of” means division by 3.
Answer:
2, 5
Step-by-step explanation:
10 = 2 × 5 . . . . . . both of these factors are prime
(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
