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Harrizon [31]
3 years ago
12

Can all linear functions be modeled by an arithmetic sequence? if not, provide a counterexample

Mathematics
1 answer:
Brut [27]3 years ago
4 0
It depends. Generally no.

Linear equations are generally in the form [math]y=mx+b[/math] and have a domain of [math](-\infty,\infty)[/math], or all real numbers. However, an arithmetic sequence is only defines for the natural numbers (that is, while numbers [math]> 0[/math].

For any two terms in the arithmetic sequence, [math]a_n[/math] and [math]a_{n+1}[/math], there will always be a point on the linear function that lies in between them, and is such not defined in the sequence.

This does not make the sequence and function unrelated, but rather it makes them not the same.

A similar argument applies for geometric sequences and exponential equations.
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Help!! Don’t really understand this!!
tekilochka [14]

Answer:

15

Step-by-step explanation:

You find the square root of 5 and 3 then square the answers you get which just undos the square root so in the end you just have 5 times 3 which is 15.

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3 years ago
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Roast Beef: 32
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3 years ago
What is the value of the expression, written in standard form?
Studentka2010 [4]

Answer:

The value of the expression = 200

Step-by-step explanation:

The given expression  is \frac{6.6 * 10^{-2}}{3.3*10^{-4}}

Using the low of exponent x^{a} /x^{b}  = x^{a-b}

So, the expression  =

\frac{6.6 * 10^{-2}}{3.3*10^{-4}} = \frac{6.6}{3.3} * 10^{-2 -(-4)} = 2 * 10^{2} = 2*100=200

<u>So, The value of the expression = 200</u>

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3 years ago
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Which equation is correct for the perpendicular bisector of the line segment whose endpoints are (−1,1) and (7,−5)?
kherson [118]
Slope = (-5-1)/(7+1) = -6/8 = -3/4
Perpendicular slope = 4/3

Midpoint = ((-1+7)/2 , (1+(-5))/2) = (3, 3)

equation:
y = mx + c
y = 4/3 x + c

at point (3, 3)
3 = 4/3(3) + c
c = 3 - 4 = -1

y = 4/3 x - 1 or 3y = 4x - 3

6 0
3 years ago
Gabriel deposits $2,500 into each of two savings accounts.
charle [14.2K]

Answer:

$5,612.16

Step-by-step explanation:

Part 1) Account I earns 4% annual simple interest.

we know that

The simple interest formula is equal to

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

substitute in the formula above

Part 2) Account II earns 4% interest compounded annually.

we know that    

The compound interest formula is equal to  

A=P(1+rt)

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=3 years

p=$2,500

r=4%=4/100=0.04

substitute in the formula above  

A=2,500(1+0.04 * 3)

A=2,500 (1.12)

A=$2,800

Part 3) What is the sum of the balances of Account I and Account II at the end of 3 years?

Sum the two final investment

$2,800 + $2,812.16 = $5.612.16

8 0
4 years ago
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