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3 years ago

Write an explicit formula for the sequence given by the recursive definition a(1)=1 and a(n+1)=a(n)+7

Mathematics

1 answer:

Basile [38]3 years ago

5 0

Answer:

an = -6+n

an = 1 + 7(n-1)

Step-by-step explanation:

a(1)=1 and a(n+1)=a(n)+7

The explicit formula is

an = a1+ d (n-1)

we know a1 =1

Looking at a(n+1)=a(n)+7

We are adding 7 each time so the common difference is +7

an = 1 + 7(n-1)

We can simplify this

an = 1 + 7n -7

an = -6+n

You can use either formula

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Ssimpily the expression

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3 years ago

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riadik2000 [5.3K]

Answer:

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Step-by-step explanation:

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3 years ago

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How do you solve for -7.21/-3.5

AVprozaik [17]

Answer:

it's 2.06

Step-by-step explanation:

$\frac{ - 7.21}{ - 3.5} = 2.06$

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3 years ago

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seraphim [82]

Answer:

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Step-by-step explanation:

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3 years ago

The tangent of the graph of y=<img src="https://tex.z-dn.net/?f=e%5E%7Bax%7D" id="TexFormula1" title="e^{ax}" alt="e^{ax}" align

bearhunter [10]

a) The tangent to $y=e^{ax}$ at $x=p$ has slope

$y' = ae^{ax} \implies y'(p) = ae^{ap}$

and $y=e^{ap}$ at this point. It passes through the origin, so its equation is

$y - 0 = ae^{ap} (x - 0) \implies y = ae^{ap} x$

It also passes through the point $(p,e^{ap})$ on the curve, so

$y - e^{ap} = ae^{ap} (x - p) \implies y = e^{ap} + ae^{ap} x - ape^{ap}$

By substitution,

$ae^{ap} x = e^{ap} + ae^{ap} x - ape^{ap} \implies e^{ap} = ape^{ap} \implies ap=1 \\\\ \implies \boxed{p=\dfrac1a}$

b) The normal to $y=e^{ax}$ at $x=2p$ has slope

$-\dfrac1{y'(2p)} = -\dfrac1{ae^{2ap}} = -1 \implies ae^{2ap} = 1$

It follows that

$ae^{2ap} = ae^2 = 1 \implies \boxed{a = \dfrac1{e^2} \text{ and } p = e^2}$

c) The tangent line equation is then

$y = \dfrac1{e^2} e^1 x \implies \boxed{y = \dfrac xe}$

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2 years ago