P: Parenthesis
E: Equation
M:Multiplication
D: Division
A : addition
S: Subtraction
So first you need to divide because it is before so you divide 18 by 6 which is 3 then you subtract (31 - 3) which gives you 28
answer : 28
Given that

, then

The slope of a tangent line in the polar coordinate is given by:

Thus, we have:

Part A:
For horizontal tangent lines, m = 0.
Thus, we have:

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
</span><span>θ = 0
</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488
Part B:
For vertical tangent lines,

Thus, we have:

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
</span>θ = <span>4.91718592528713</span>
Answer:
Step-by-step explanation:
1 2/3 (5 2/7)
For this problem, the confidence interval is the one we are looking
for. Since the confidence level is not given, we assume that it is 95%.
The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n
Where:
<span>
</span>
α= 5%
α/2
= 2.5%
t
0.025, 19 = 2.093 (check t table)
n
= 20
df
= n – 1 = 20 – 1 = 19
So plugging in our values:
8.41 ± 2.093 * 0.77 √ 1 + 1/20
= 8.41 ± 2.093 * 0.77 (1.0247)
= 8.41 ± 2.093 * 0.789019
= 8.41 ± 1.65141676
<span>= 6.7586 < x < 10.0614</span>
Hi the answer to this is
A - Vertical Asymptote