Answer:
θ = 49.4584°
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Trigonometry</u>
- [Right Triangles Only] SOHCAHTOA
- [Right Triangles Only] cosθ = adjacent over hypotenuse
- Inverse Trig
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
Angle θ = ?
Adjacent Leg = 13
Hypotenuse = 20
<u>Step 2: Find Angle</u>
- Substitute in variables [cosine]: cosθ = 13/20
- Inverse Trig: θ = cos⁻¹(13/20)
- Evaluate: θ = 49.4584°
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=6The arclength of a parametric curve can be found using the formula:
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=
6
√
3
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Answer:
Step-by-step explanation:
What is it?
The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
How do you find IQR?
<em>Step 1: Put the numbers in order. ...</em>
<em>Step 2: Find the median. ...</em>
<em>Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...</em>
<em>Step 4: Find Q1 and Q3. ...</em>
<em>Step 5: Subtract Q1 from Q3 to find the interquartile range.</em>
