Answer:
½ sec²(x) + ln(|cos(x)|) + C
Step-by-step explanation:
∫ tan³(x) dx
∫ tan²(x) tan(x) dx
∫ (sec²(x) − 1) tan(x) dx
∫ (sec²(x) tan(x) − tan(x)) dx
∫ sec²(x) tan(x) dx − ∫ tan(x) dx
For the first integral, if u = sec(x), then du = sec(x) tan(x) dx.
∫ u du = ½ u² + C
Substituting back:
½ sec²(x) + C
For the second integral, tan(x) = sin(x) / cos(x). If u = cos(x), then du = -sin(x) dx.
∫ -du / u = -ln(u) + C
Substituting back:
-ln(|cos(x)|) + C
Therefore, the total integral is:
½ sec²(x) + ln(|cos(x)|) + C
Answer:seahawks | 20
sharks | 14
marlins. | 12
dolphins. | 9
penguins | 5
Step-by-step explanation:
The <u>correct answer</u> is:
<span>Relative frequencies are the probabilities occurring in sampling distributions.
Explanation:
Relative frequencies are the fraction of times an event occurs within a sample.
This is the same definition as experimental probability; thus relative frequencies are the probabilities occurring in sampling distributions.</span>
Answer:
251.33in²
Step-by-step explanation:
A cylindrical container contains some sand. If the diameter of the container is 10 inches and its height is 3 inches, about how much sand fits inside the container?
We solve the above question using the formula for Total surface area of a cylinder.
TSA = 2πr (h + r) Square units.
Where
h = Height of the cylinder
r = Radius of the cylinder
Diameter of the container is 10 inches
Hence, radius = Diameter/2 = 10/2 = 5 inches
Height is 3 inches.
Hence, Total surface area =
2 × π × 5 (5 + 3)
= 10π × (8)
= 251.32741229 in²
Approximately = 251.33 in²
Therefore, the amount of sand that can fit into the cylinder = 251.33 in²
