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2 years ago

If x an integer and 2x +8 <22 and 3x >9, how many possible values of x are there?

Mathematics

2 answers:

11111nata11111 [884]2 years ago

6 0

Answer:

4, 5 and 6.

Step-by-step explanation:

2x + 8 < 22

2x < 14

x < 7.

3x > 9

x > 3.

So the possible values if x is an integer are 4, 5 and 6.

koban [17]2 years ago

5 0

Answer:

3 values

Step-by-step explanation:

2x +8 <22 and 3x >9

Solve the two inequalities separately and then put them back together

2x +8 <22

Subtract 8 from each side

2x +8-8 <22 -8

2x < 14

Divide by 2

2x/2 < 14/2

x < 7

Then solve the second inequality

3x>9

Divide by 3

3x/3>9/3

x >3

Putting them back together

x>3 and x < 7

Since it has to be an integer

4,5,6 are possible choices

3 values

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2 years ago

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We need to simplify the given expression.

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3 years ago

What 2 numbers multiply to be 108 but subtract to be 6

mel-nik [20]

No 2 numbers do this.

Possible numbers that can multiply to 108:

<span>1 x 108 = 108
2 x 54 = 108
3 x 36 = 108
4 x 27 = 108
6 x 18 = 108
9 x 12 = 108
12 x 9 = 108
18 x 6 = 108
27 x 4 = 108
36 x 3 = 108
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Possible numbers on that list that when subtracted equate to 6:

None</span>

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2 years ago