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Anton [14]
3 years ago
12

Help me please i will follow

Mathematics
2 answers:
alukav5142 [94]3 years ago
5 0

Answer:

probably not the equation your looking for but 2+4x, 2+4(7) = 30

Tcecarenko [31]3 years ago
4 0

Answer:

think it's c

Step-by-step explanation:

two people on each side of the table will be 2 so on 7 table it will be 14 on one side and 14 on the other, that is 28. then you need to add the ends so 2. so now add 28 and 2 and that is 30

You might be interested in
Given the functions f(x) = 3(0.56)^x and g(x) = 6(4)^x
rodikova [14]
A)  f(x) is decreasing because the base is less than 1.
      0.56 is close to 0.5, so its like saying that you are taking half each time, therefore the value is getting smaller.

g(x) is increasing because the base is greater than 1.
you are multiplying by 4 each time, making the value bigger.


B )  The y-intercept is where x=0.
      Anything to the '0' power is 1. Therefore the y-intercept is equal to the coefficient in front of each function.
    f(x) = 3 ,   g(x) = 6


C)  Just plug in x=4 to each function in a calculator.
 f(4) = 0.295
g(4) = 1536
7 0
3 years ago
The Eurasian tectonic plate travels at a
Ronch [10]

Answer:

17.5 cm

Step-by-step explanation:

Multiply 0.7 by 25 to get 17.5 cm.

Hope it helped.

8 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
Help would be very needed
Ipatiy [6.2K]

Answer:

Step-by-step explanati

28 cm2

4 0
3 years ago
Midpoint of the line segment (-1,6), (-6,5)
zepelin [54]
The answer is:

(-7/2 , 11/2)

5 0
2 years ago
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