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Olegator [25]
3 years ago
8

Graph the set of points. Which model is most appropriate for the set?

Mathematics
1 answer:
Mila [183]3 years ago
4 0

Answer:

A. quadratic

Step-by-step explanation:

the four points forms a curve that is parallel (parabola). quadratic is the shape of a curve like that

and exponential is only a curve

linear is like a straight line

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What is the answer of Three over <br> Four multiply to eight over nine ?
andrew11 [14]
The answer would be 2 over 3, or 2/3. (:
5 0
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PLEASE PLEASE PLEASE HELP<br> WILL GIVE BRAINLIEST<br><br> 2/5 x + -6 + 7/15 x
Klio2033 [76]

Answer:

13x/15-6

Step-by-step explanation:

i used symlobab

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In one area, the lowest angle of elevation of the sun in winter is 27.5°. Find the minimum distance x that a plant needing full
Alex73 [517]

Answer:

The minimum distance x that a plant needing full sun can be placed from a fence that is 5 feet high is 4.435 ft

Step-by-step explanation:

Here we have the lowest angle of elevation of the sun given as 27.5° and the height of the fence is 5 feet.

We will then find the position to place the plant where the suns rays can get to the base of the plant

Note that the fence is in between the sun and the plant, therefore we have

Height of fence = 5 ft.

Angle of location x from the fence = lowest angle of elevation of the sun, θ

This forms a right angled triangle with the fence as the height and the location of the plant as the base

Therefore, the length of the base is given as

Height × cos θ

= 5 ft × cos 27.5° = 4.435 ft

The plant should be placed at a location x = 4.435 ft from the fence.

8 0
3 years ago
Harry church borrowed 17,500.00 at 6.5% exact interest. He had to pay back a maturity value of 17,873.97 to pay off the loan. Wh
kirill [66]

Answer:the term of the loan is approximately 4 months

Step-by-step explanation:

The term of the loan means the period for which the loan was given.

We would apply the formula for simple interest which is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

From the information given

P = 17500

R = 6.5%

I = total amount paid - principal

I = 17,873.97 - 17,500.00 = 373.97

Therefore

373.97 = (17500 × 6.5 × T)/100

373.97 = 1137.5T

T = 373.97/1137.5

T = 0.32 years

Converting to months, it becomes

0.32 × 12 = 3.84

Approximately 4 months.

3 0
3 years ago
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
3 years ago
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