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madam [21]
2 years ago
15

Write an equation that expresses the following relationship.

Mathematics
1 answer:
devlian [24]2 years ago
5 0

Answer:

Step-by-step explanation:

p = kd/u

p varies directly with d, p=md

p varies inversely with u, p=n/u

put them together

p=kd/u

as d goes up, p goes up.  as u goes up, p goes down

You might be interested in
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
2 years ago
Solve for X <br> If you have 2x+3
daser333 [38]

Answer:

in the explanation

Step-by-step explanation:

According to the triangle midsegment theorem (if these points are midpoints),

2x + 3 is two times longer than 7, meaning it's 14.

2x+3=14

2x=11

x = 11/2

If there is an option for not enough info given, that should be right as it doesn't actually say that they are midpoints

5 0
3 years ago
Please help ill give brainiest answer
trapecia [35]

Answer:

1: AAS, RQC 2: ASA, SRP

Step-by-step explanation:

(1) We are shown that two angles and one side are congruent in the order of AAS. Make sure you write the letters in terms of the corresponding angles. The angles and sides are congruent because the problem labels it for us. For example, B,A,C=R,Q,S. Answers: AAS, RQC.

(2) We are shown that two angles and one side are congruent in the order of ASA. Make sure you write the letters in terms of the corresponding angles again. For example, P,Q,R=P,S,R. The angles are congruent because the problem labels it for us. Side PR is congruent to side PR by reflexive property. Answers: ASA, SRP.

I hope this helped :) Good luck

4 0
3 years ago
How can you tell when a pattern shows counting on by tens
FromTheMoon [43]
All the number end in a zero 10 20 30 40 50 60 70 80 90
4 0
3 years ago
taro has marbles whose total mass is 24g if each marble has a mass of about 1.5g how many marbles does taro have
guajiro [1.7K]
24 divided by 1.5 equals 16

Taro has 16 marbles
3 0
2 years ago
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