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adoni [48]
1 year ago
6

Which of the following is a Recursive Formula for an Arithmetic Sequence?

Mathematics
1 answer:
In-s [12.5K]1 year ago
3 0

The Recursive formula for an Arithmetic Sequence is a1 = -3, an = an-1 + 6.

<h3>What is Arithmetic sequence ?</h3>

An arithmetic sequence is a sequence where each term increases by adding/subtracting some constant k.

Given:

an = -3 + 6(n – 1)

a1= -3

a2= 3

a3 = 9

Here the common difference is 6 and first term is -3

Hence, the recursive formula is a1 = -3, an = an-1 + 6.

Learn more about this concept here:

brainly.com/question/12942847

#SPJ1

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Choose the equation that represents the line passing through the point -3,-1 with a slope of 4
lubasha [3.4K]

Answer:y=4x+11

Step-by-step explanation:

Substitute the coordinates in each of the above equations,

Then the left hand side of the equation should be equal to the right hand side of the equation.

By substituting the above coordinates in the second answer,

-1= 4(-3)+11

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6 0
3 years ago
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T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


{y}''=7e^{x}\\ {y}'=7e^{x}


k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

7(e^x)(1+49e^{2x})(98e^{2x}-1)=0

→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

e^{x}=0  ,  ∧ 1+49e^{2x}=0   [not possible ∵there exists no value of x satisfying these equation]

→98e^{2x}-1=0

Solving this we get

x= -\frac{1}{2}\ln{98}

As you will evaluate {k(x})}''<0 at x=-\frac{1}{2}\ln98

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[-\frac{1}{2}\ln98,1/√2]

k(x)=\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}

k(x)=\frac{7}{\infty}

k(x)=0







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