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Lelu [443]
3 years ago
7

Add 3x to both sides

Mathematics
2 answers:
cricket20 [7]3 years ago
6 0
To both sides of what?
ollegr [7]3 years ago
5 0

Answer:

Add 3x to both side of what? please write the whole question.

Step-by-step explanation:

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QUIZ
oksian1 [2.3K]

Answer:

You need to include the graph

Step-by-step explanation:

You need to include the graph, you just pasted everything you copied.

6 0
4 years ago
Which is an equivalent form of the expression 1800 (1.2) ^-t
ziro4ka [17]
I think this is how you do it. Hope this helps!!

4 0
3 years ago
Read 2 more answers
when 4 is added to the difference between 5 times a number and 9 the result is greater than 10 * to 4 times the number what's th
Lubov Fominskaja [6]
5x - 9 > x*10^4
5x - 9 > x * 10000 I hope I'm reading the question correctly.
5x > 10000 x + 9
x > 10000/5 x + 9/5
x > 2000x + 1.8
0 > 1999x + 1.8
- 1.8 > 1999x
- 1.8 / 1999 > x
- 0.0009004 > x
3 0
3 years ago
can the measure of an angle ever equal exactly half the measure of the supplement of an angle (please answer question correctly)
Leni [432]

Answer:

Let the angle be x.

Its complement is 90°-x° and its supplement is 180°- x.

If 90°-x = (1/2)(180°-x) then 180°- 2x = 180° -x => 180°-180° = 2x-x => x= 0°

Hence the statement is true only for 0°

8 0
3 years ago
An arithmetic sequence has this recursive formula:
maxonik [38]

An arithmetic sequence can be expressed as explicitly or recursively

The explicit formula of the sequence is a_n = 11 - 4n

<h3>How to determine the explicit formula</h3>

The recursive formula is given as:

a_n = a_{n -1} - 4

a_1 = 7

Substitute 2 for n in a_n = a_{n -1} - 4

a_2 =a_1 - 4

This gives

a_2 =7 - 4

a_2 =3

Calculate the common difference (d)

d = a_2 -a_1

d = 3- 7

d = -4

The explicit formula is then calculated as:

a_n = a_1 + (n - 1)d

This gives

a_n = 7+ (n - 1)*-4

Expand

a_n = 7+4 - 4n

a_n = 11 - 4n

Hence, the explicit formula of the sequence is a_n = 11 - 4n

Read more about arithmetic sequence at:

brainly.com/question/6561461

5 0
3 years ago
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