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olya-2409 [2.1K]
3 years ago
5

What fomula can be used to describe the sequence? -2/3,-4,-24,-144

Mathematics
1 answer:
Lunna [17]3 years ago
5 0

This is a geometric sequence with n th term formula

a_{n} = a_{1}(r)^{n-1}

where r is the common ratio and a_{1} the first term

here a_{1} = - \frac{2}{3} and

r = \frac{-144}{-24} = \frac{-24}{-4} = 6

a_{n} = - \frac{2}{3}(6)^{n-1}

or in terms of x

f(x) = - \frac{2}{3}(6)^{x-1}



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SpyIntel [72]

Answer:

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Step-by-step explanation:

IF you are asking why it is because 50-32=18 than you do 18/.90 which equals 20 so there you go thanks make brainliest

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What is <br><br> 3x+2y=-13<br> 3x+4y=1
Zepler [3.9K]

Answer:

The values are

x = -25/9 = -2 7/9

y = 7/3 = 2 1/3

Step-by-step explanation:

3x + 2y = -13 --------eqn 1

3x + 4y = 1-------------eqn2

Using eqn 2 to get the value of y

3x + 4y = 1

4y = 1 - 3x

Dividing both sides by 4,to get y

4y/4 =( 1 -3x) / 4

y = (1 - 3x) / 4

Since we've gotten the value for y, substitute the value into eqn 1

3x + 2y = -13

3x + 2(3x - 1)/4 = -13

Opening the bracket

3x + (6x - 2)/4 = -13

LCM = 4

(12x + 6x - 2) / 4 = -13

18x - 2 / 4 = -13

Then we cross multiply

18x - 2 = -13 * 4

18x - 2 = - 52

18x = -52 + 2

18x = -50

Divide both sides by 18, to get the value of x

18x/18 = -50/18

x = -25/9

or x = -2 7/9

The value of x is now known, so let's go back to eqn 2

Substitute x = - 25/9

3x + 4y = 1

3(-25/9) + 4y = 1

Open the bracket

-75/9 + 4y = 1

Make y the subject of the formula

4y = 1 + 75/9

LCM = 9

4y = (9 + 75)/ 9

4y = 84/9

To get y, divide both sides by 4

4y/4 = 84/9 / 4/1

y =

Note : when division changes to multiplication, it always be in its reciprocal form

y = 84/9 / 1/4

y = 84 * 1 / 9 *4

y = 84/ 36

y = 7/3

Or

y = 2 1/3

4 0
3 years ago
VI
barxatty [35]

Answer:D hope it helped

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Simplify the scalar multiplication on the matrix.
Romashka-Z-Leto [24]

Answer:

third option

Step-by-step explanation:

Given

3 \left[\begin{array}{ccc}-2&5\\1&0\\\end{array}\right]

Multiply each element in the matrix by 3

= \left[\begin{array}{ccc}3(-2)&3(5)\\3(1)&3(0)\\\end{array}\right]

= \left[\begin{array}{ccc}-6&15\\3&0\\\end{array}\right]  

7 0
3 years ago
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