Can yoy please help solve yhis problem.2/5b+1=11
2 answers:
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1)
v - 6 ≥ 4
<u> +6 </u> <u>+6 </u>
v ≥ 10
Graph: 10 -----------------→ <em>the dot at 10 is filled in because of the "equal to" symbol</em>
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2)
-5x < 15
<em>divided by a negative so the symbol flipped </em>
x > -3
Graph: -3 o------------------→ <em>the dot at -3 is NOT filled in because it's NOT "equal to" </em>
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3)
3k > 5k + 12
<u>-5k </u> <u>-5k </u>
-2k > 12
<em>divided by a negative so the symbol flipped </em>
k < -6
Graph: ←------------o -6 <em>the dot at -6 is NOT filled in because it's NOT "equal to" </em>
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5)
2t ≤ 4 or 7t ≥ 49
or
t ≤ 2 or t ≥ 7
Graph: ←------- 2 7 ---------→ <em>the dots at 2 and 7 are filled in</em>
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6)
| n + 2 | = 4
n + 2 = 4 or n + 2 = -4
<u> -2</u> <u>-2 </u> <u> -2 </u> <u>-2 </u>
n = 2 or n = -6
Answer: n = {2, -6}
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7)
| 2x - 7 | > 1
2x - 7 > 1 or 2x - 7 < -1
<u> +7</u> <u>+7 </u> <u> +7 </u> <u>+7 </u>
2x > 8 or 2x < 6
or
x > 4 or x < 3
Graph: ←----------o 3 4 o------------→
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8)
A = {1, 2, 3, 4, 5, 6, 7, 8, 9} B = {2, 4, 6, 8}
A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9} <em>union combines both sets</em>
A ∩ B = {2, 4, 6, 8} <em>intersection includes only those that are in BOTH sets</em>
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9)
P = {1, 5, 7, 9, 13} R = { 1, 2, 3, 4, 5, 6, 7} Q = {1, 3, 5}
P ∩ R ∩ Q = {1, 5}
P ∩ R = {7} <em>disregard 1 & 5 since they are already in </em>P∩Q∩R
R ∩ Q = {3} <em>disregard 1 & 5 since they are already in </em>P∩Q∩R
P ∩ Q = { } <em>disregard 1 & 5 since they are already in </em>P∩Q∩R
P no intersection = {9, 13)
R no intersection = {2, 4, 6}
Q no intersection = { }
If you can't figure out how to draw it based on the information I provided, please see the attached diagram. <em>Note: Usially, you would only identify P, Q, R. I identified the intersections so you would understand why those specific numbers are placed in the designated sections.</em>
