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faust18 [17]
3 years ago
8

A geometric sequence is defined recursively by an = 20an-1. The first term of the sequence is 0.01. Which of the following is th

e explicit formula for the nth term of the sequence?
A. 0.01*20n+1
B. 0.01*20n
C. 20*0.01n-1
D. 0.01*20n-1
E. 20 *0.01n
Mathematics
1 answer:
Dmitrij [34]3 years ago
6 0
Hello : 
hello :
the nth term is : An=A1×r^(n-1) 
<span>The first term of the sequence is 0.01
r =An/an-1 =20
</span><span>An =0.01(20)^n-1...(answer :D )
</span>
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Three consecutive integers add to 567. Find the integers. PLEASE HELP
Aleks04 [339]
Consecutive integers mean one after the other and their difference would one more or less.

Let the numbers be x, x +1, x +2.

Therefore x + (x+1) + (x +2) = 567
               3x + 3 = 567
               3x = 567 -3 = 564
               3x = 564  Divide by 3.
                 x  = 564/3
                 x  = 188
Therefore integers  x, x +1, x +2 = 188, 188+1, 188+2.
                                                =  188, 189, 190.
Cheers.
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Could anyone help me with this ?!
allochka39001 [22]

Answer: 5; PO

6; PM

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Step-by-step explanation:

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=5x-%20%5Cdfrac%7B%20%5Csqrt%7B%209%20%7B%20x%20%7D%5E%7B%202%20%7D%20-6x%2B1%20%5Cphantom%7B%5
Morgarella [4.7K]

Answer:

=5x+1          or          =5x-1

Step-by-step explanation:

One is given the following equation:

5x-\frac{\sqrt{9x^2-6x+1}}{1-3x}

The problem asks one to simplify the expression, the first step in solving this equation is to factor the equation.  Rewrite the numerator and denominator of the fraction as the product of two expressions. Remember the factoring patterns:

(a-b)^2=a^2-2ab+b^2

=5x-\frac{\sqrt{9x^2-6x+1}}{1-3x}

=5x-\frac{\sqrt{(3x-1)^2}}{-(3x-1)}

Now simplify the numerator. Remember, taking the square root of a squared value is the same as taking the absolute value of the expression,

=5x-\frac{\sqrt{(3x-1)^2}}{1-3x}

=5x-\frac{|3x-1|}{-(3x-1)}

Rewrite the expression without the absolute value sign in the numerator. Remember the general rule for removing the absolute value sign:

|a-b|\\=a -b           or           (-a-b) = b-a

=5x-\frac{|3x-1|}{-(3x-1)}

=5x-\frac{3x-1}{-(3x-1)}          or          =5x-\frac{-(3x-1)}{-(3x-1)}

Simplify both expressions, reduce by canceling out common terms in both the numerator and the denominator,

=5x-\frac{3x-1}{-(3x-1)}          or          =5x-\frac{-(3x-1)}{-(3x-1)}

=5x-(-1)          or          =5x-(1)

Simplify further by rewriting the expression without the parenthesis, remember to distribute the sign outside the parenthesis by the terms inside of the parenthesis; note that negative times negative equals positive.

=5x-(-1)          or          =5x-(1)

=5x+1          or          =5x-1

6 0
3 years ago
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