3. <span>Suppose U = {1, 2, 3, 4, 5} is the universal set and A = {2, 3}. What is A ? Well "A" is {2, 3}, but I'm guessing you meant A'. A' is all the numbers you don't see, which is {1, 4, 5}
4. </span><span>Suppose U={1, 2, 3, 4, 5, 6, 7, 8} is the universal set and P={2,4, 6, 8}. What is P? Again, "P" is just {2,4, 6, 8}, but P' is all the numbers you don't see, which are all prime numbers in this sequence: {1, 3, 5, 7}
5. </span><span>-4 < k + 3 < 8 subtract 3 from all sides -7 < k < 5
6. </span><span> 5 <= y + 2 <= 11 subtract 2 from all sides 3 <= y <= 9
7. </span><span>6b - 1 < -7 or 2b + 1 > 5 solve both </span>6b - 1 < -7 add both sides by 1 6b < -6 divide both sides by 6 b < -1
now do the other problem 2b + 1 > 5 subtract both sides by 1 2b > 4 divide both sides by 2 b > 2 answer: b < -1 or b > 2
<span>8. 5 + m > 4 or 7m < -35 subtract both sides by 5 |or| divide both sides by 7 m > -5 or m < -5 m = -5</span>
The ratio in which Aliyah, Brenda and Candy share the sum of money = 3:5:6
The amount Candy later gives Aliyah = $100
The amount Candy later gives Brenda = $50
The new ratio of the sum of the shared money between Aliyah, Brenda and Candy = 2:3:3
(a) Whereby Aliyah has $3x at the start, we have;
Total sum of mony = Y
Amount of Aliyah's initial share = Y × 3/(3 + 5 + 6) = Y×3/14
Therefore, Y×3/14 = $3x
x = Y×3/14 ÷ 3 = Y/14
Amount of Candy's initial share = Y × 6/14
Therefore Candy's initial sum as a terms of x = $6x
(b) Given that Aliyah's and Candy's initial sum as a function of x are $3x and $6x, therefore, in the ratio 3:5:6, Brenda's initial sum as a function of x = $5x
Which gives;
Total amount of money = $14x
With
6x - 150, 3x + 100, and 5x + 50, the ratio =is 2:3:3
Therefore, we have;
14·x × 2/(2 + 3 + 3) = (6·x - 150)
14·x × 2/(8) = (6·x - 150)
14·x × 1/4 = (6·x - 150)
7·x/2 = (6·x - 150)
12·x - 300 = 7·x
12·x - 7·x = 300
5·x = 300
x = $60
(b) The final amount of money with Brenda = 5x + 50 = 5 × 60 + 50 = $350
