Michelle kicks a soccer ball up into the air from the ground. The graph represents the height of the ball, in feet, as a functio
n of time, in seconds. image 84feafed15d34870b0898ecea9d9cf41 Select all of the values that are in the domain of the function represented by the graph. A -5 seconds B 0 seconds C 2.5 seconds D 6 seconds E 10 seconds
2 answers:
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Answer:
2nd debate was 3 hours long.
Step-by-step explanation:
We have been given that during the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours. The second one was 1 4/5 times as long as the first one.
Let us find the estimate of time spent on 2nd debate.
1 2/3 hours would be approximately 2 hours. 1 4/5 times would be equal to 2 times.
Therefore, the estimated time is less 4 hours.
To find the time spent on 2nd debate, we will multiply 1 2/3 by 1 4/5.
First of all, we will convert mixed fractions into improper fractions as:
Now, we will multiply both fractions as:
Therefore, the 2nd debate was 3 hours long.
Step-by-step explanation:
In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
