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3 years ago

If the nth term of a sequence is 28-3n what is the 7th term?

2 answers:

5 0

Erms.
For 3, 9, 27, ... the common ratio is 3 because:
3 X 3 = 9
9 X 3 = 27
So to find the 7th term you can do it two ways:
One way:
3 is the 1st term, 9 is the 2nd term, 27 is the 3rd term so then
4th term: 27 X 3 = 81
5th term: 81 X 3 = 243
6th term: 243 X 3 = 729
7th term: 729 X 3 = 2,187
Another way:
You can use the explicit formula
a
n
=
a
1
⋅
r
n
−
1
, where
a
n
is the nth term,
a
1
is the first term, n is the number of the term, and r is the common ratio
so
a
7
=
3
⋅
3
7
−
1

a
7
=
3
⋅
3
6

a
7
=
3
⋅
729

a
7
=
2
,
187

Both ways get you to the same answer that the 7th term in that geometric sequence is 2, 187 .

expeople1 [14]3 years ago

3 0

Answer:

Step-by-step explanation:

28-3(7)

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2 years ago

Use the scenario to answer the question

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Answer:

-$5.27 per day

Step-by-step explanation:

1st. $300

6th. $273.65

19th. $205.14

25th. $173.52

28th. $157.71

Lets use the one that has the least amount of days separated to make it easier

28th and 25th

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2 years ago

A triangle weighs 3 grams and a circle weighs 6 grams. Find the weight of a square in Hanger A and the weight of a pentagon in H

Zanzabum

Answer:

Equation of Hanger A:

x + y = z

z = 9 grams

Equation for hanger B

x + y = p

p = 9 grams

Step-by-step explanation:

Note: This question is not complete, and lacks necessary data to answer this question. But for the sack of concept, we wil be solving this question using our own data.

Data given:

Triangle weight = 3 grams

Circle weight = 6 grams

There are two hangers A and B.

We have to find weight of the square in Hanger A and weight of the pentagon in Hanger B.

So, for your understanding I have drawn the two hangers A and B. In A we have triangle and circle on the LHS and the square on the right. On the other hand, we have pentagon on the right and triangle and circle.

Sketch is attached in the attachment.

Let's suppose: Hanger A is balanced, So,

Triangle weight = 3 grams = x

Circle weight = 6 grams = y

weight of the pentagon in Hanger B = p

weight of the square in Hanger A = z =?

So,

Equation of Hanger A:

x + y = z

z = 3 + 6

z = 9 grams = weight of the square in Hanger A

Similarly for Hanger B.

Equation for hanger B

x + y = p

p = 3 + 6

p = 9 grams

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2 years ago