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Eduardwww [97]
3 years ago
13

Write a recursive rule and explicit rule for the arithmetic sequence 19,9,-1,-11, ...

Mathematics
1 answer:
serg [7]3 years ago
6 0
★ Arithmetic Progression [ Explicit and Recursive notation ] ★

Arithmetic Sequence - 19 , 9 , -1 ...

Recursive Notation -

First term = 19
Common difference = -10

In subscript notation - [ a₁ = 19 , aₙ = aₙ-₁ -10 ]
In functional notation - f(n) = f (n -1) -10

Explicit Notation - f ( n ) = f ( 1 ) + d ( n - 1 )

= 19 - 10 ( n -1 )
= 29 - 10n

It's functional and subscript notations resides the same


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Eli has saved $48.50. Eli's brother borrowed 1/10 of Eli's savings. How much money does Eli have left?
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$43.65

Step-by-step explanation:

1. divide 48.50 by 10

4.85

2. subtract 4.85 from 48.50

3. your answer would be $43.65

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In algebra how do i convert y-7=-2/3(x+1) into standard form and what would it be?
Art [367]

Answer:

C

Step-by-step explanation:

Where, if at all possible,

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3 years ago
In the number 15.2201 how does the value of 2 in the tenths place compare to the value of the 2 in the hundredths place
lakkis [162]

Answer:

The 2 in the hundreths is 10% of the 2 in the tenths

Step-by-step explanation:

We are given the number 15.2201

#The value of 2 in the tenths:

0.2

#The value of 2 in the hundredths:

0.02

The difference of the 2 in the tenths and in the hundredths is:

=2_t-2_h\\\\=0.2-0.02\\\\=0.18\\\\\#2_h \ as \ a \ \% \ of \ 2_t\\=\frac{2_h}{2_t}\times 100\%\\\\=\frac{0.02}{0.2}\times100\%\\\\2_h=10\% \ of\  2_t

Hence, the difference between the 2 in the tenths  minus 2 in the hundreths is 0.18, 2 in the hundreths is 10% the 2 in the tenths.

6 0
3 years ago
Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(4​,5​), ​(-6​,-6​), and ​(-14​,2​
Lilit [14]

Answer:

  (-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

  (x -4)^2 +(y -5)^2 = 10^2

  x^2 -8x +16 +y^2 -10y +25 = 100

  x^2 +y^2 -8x -10y = 59

__

The epicenter is 5 units from Y, so lies on the circle ...

  (x +6)^2 +(y +6)^2 = 5^2

  x^2 +12x +36 +y^2 +12y +36 = 25

  x^2 +y^2 +12x +12y = -47

__

The epicenter is 13 units from Z, so lies on the circle ...

  (x +14)^2 +(y -2)^2 = 13^2

  x^2 +28x +196 +y^2 -4y +4 = 169

  x^2 +y^2 +28x -4y = -31

__

Subtracting the second equation from each of the other two, we get ...

  (x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

  -20x -22y = 106 . . . . eq1 -eq2

  (x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

  16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

  x = y+1 . . . . . from eq3 -eq2 divided by 16

  10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

  21y = -63 . . . . . . . . . . . . simplify, subtract 10

  y = -3

  x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).

8 0
3 years ago
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