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tia_tia [17]
3 years ago
14

Given the function f(x) =

Mathematics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

The value of f(1) is smaller than the value of f(3)

Step-by-step explanation:

<u><em>The correct question is</em></u>

Given the function f(x) = 2x^2 + 3x + 10, find f(1) and f(3). Choose the statement that is true concerning these two values.

The value of f(1) is the same as the value of f(3).

The value of f(1) cannot be compared to the value of f(3).

The value of f(1) is larger than the value of f(3).

The value of f(1) is smaller than the value of f(3)

we have

f(x)=2x^{2}+3x+10

step 1

Find out the value of f(1)

substitute the value of x=1 in the function f(x)

so

For x=1

f(1)=2(1)^{2}+3(1)+10

f(1)=15

step 2

Find out the value of f(3)

substitute the value of x=3 in the function f(x)

so

For x=3

f(3)=2(3)^{2}+3(3)+10

f(3)=37

step 3

Compare the values

37> 15

so

f(3) > f(1)

or

f(1) < f(3)

therefore

The value of f(1) is smaller than the value of f(3)

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3 0
2 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
2 years ago
60&gt;5(6-2x) ok help me so I can get an answer
Dima020 [189]

Inequalities are used to represent unequal expressions

The solution to the inequality is: x > -3

<h3>How to determine the inequality solution</h3>

The inequality is given as:

60>5(6-2x)

Rewrite the inequality properly as:

60>5(6-2x)

Divide both sides of the inequality by 5

12>6-2x

Subtract 6 from both sides of the inequality

6>-2x

Divide both sides of the inequality by -2

-3Rewrite the inequality as[tex]x > -3

Hence, the solution to the inequality is: x > -3

Read more about inequalities at:

brainly.com/question/11234618

5 0
2 years ago
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