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sammy [17]
3 years ago
15

Which is the graph of f(x) = 4[1/2]x ?

Mathematics
1 answer:
bagirrra123 [75]3 years ago
6 0

Step-by-step explanation:

answer is in picture see

hope it helpful

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David used the measuring cup show to measure milk for a batch of muffins. He needs to make 12 batches of muffins for a fund rais
lozanna [386]
16x12=192
Yes, if he only uses one cup per batch
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3 years ago
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-x=-15<br><br>Syep by step solution need​
andrew-mc [135]

Answer:

x = 15

Step-by-step explanation:

-x = -15

x = -1 × (-15)

<u>x</u><u> </u><u>=</u><u> </u><u>1</u><u>5</u>

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2 years ago
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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
Rearrange a (q-c)=d to make q the subject
Andrej [43]
To rearrange you must first simplify
aq-ac=d  
add ac to each side:
aq=d+ac
then divide by a:
q=(d/a)+c and this is your answer
3 0
3 years ago
Solve the following equations.<br> √3 ⋅ 3^3x = 9
GREYUIT [131]

Answer:

The solution is x = 0.5

Step-by-step explanation:

We need to write everything as a power of 3.

We know that:

\sqrt{a} = a^{0.5}

So

\sqrt{3} = 3^{0.5}

And

9 = 3^{2}

This following property is also important:

\frac{a^{b}}{a^{c}} = a^{b-c}

To solve, the first step is putting everything with the variable x on one side and everything without the variable x on the other side

\sqrt{3}.3^{3x} = 9

3^{0.5}.3^{3x} = 3^{2}

3^{3x} = \frac{3^{2}}{3^{0.5}}

3^{3x} = 3^{2-0.5}

3^{3x} = 3^{1.5}

This means that:

3x = 1.5

x = \frac{1.5}{3}

x = 0.5

5 0
3 years ago
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