If x an integer and 2x +8 <22 and 3x >9, how many possible values of x are there?
2 answers:
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.
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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<u>For -1:</u>
y = x + 6
y = 5
( -1, 5)
<u>For 0:</u>
y = x + 6
y = 6
(0, 6)
<u>For 2:</u>
y = x + 6
y = 8
(2,8)
<u>For 4:</u>
y = x + 6
y = 10
(4,10)
C=yt+y. Divide both sides by y and u get c/y=t+1 subtract 1. The ANSWER IS c/y-1=t
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Answer:
15 divided by 6 is equivalent to 5/2. When dividing that, you get 2 and then 2/4 as the decimal portion.
Simplified it would be 2 and 1/2
A + c = 158......a = 158 - c
9.90a + 5.50c = 1265
9.90(158 - c) + 5.50c = 1265
1564.20 - 9.90c + 5.50c = 1265
- 4.40c = 1265 - 1564.20
-4.40c = - 299.20
c = -299.20/-4.40
c = 68 <=== 68 child tickets sold
