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Arturiano [62]
3 years ago
8

John can reach 60 centimeters above the top of his head. If John is 2 meters tall, how high can he reach?

Mathematics
2 answers:
Mandarinka [93]3 years ago
8 0
100 centimeters in a meter, so John can reach 260 centimeters.
Sindrei [870]3 years ago
7 0
He can reach 260 centimeters, or 2.6 meters. 

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54% in simplest form
Papessa [141]
As a fraction, 54% is  27/50.
4 0
3 years ago
Read 2 more answers
(3a) You spend $72 on tapes and CDs. Each tape costs $8 and each CD costs $12.
Makovka662 [10]

Answer:

3a. 8t + 12c = 72

3b. t + c = 7

Step-by-step explanation:

number of tapes is t

number of Cds is c

so 8t + 12c = 72

and t + c = 7

so t = 7 - c

replace t = 7 - c in to the first equation

8(7 - c) + 12c = 72

56 - 8c + 12c = 72

4c = 72 - 56 = 16

c = 4

if c = 4, then t = 7 - c = 7 - 4 = 3

6 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
3 years ago
If X²+1=2x then what the value of x²?
natka813 [3]

Answer:

x² = 2x - 1

Step-by-step explanation:

x² + 1 = 2x

→ Minus 1 from both sides

x² = 2x - 1

7 0
3 years ago
Read 2 more answers
If f(x)=-x^2+3X +5 and g(x)=X^2 +2x, which graph shows the graph of (f+g)(x)?
lana [24]

Answer:

Step-by-step explanation:

+ We find the function (f+g)(x) like

(f+g)(x)= f(x) + g(x) = x^2 +3x+5 + x^2 + 2x = 2x^2 +5x + 5

+ So the graph of (f+g)(x) is a parabole with the sommet (-5/4, 15/8).

Hope that useful for you.

5 0
3 years ago
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