A) Isolate y in both inequalities
1) x + y ≥ 4 => y ≥ 4 - x
2) y < 2x - 3
B) Draw the lines for the following equalities:
1) y = 4 - x
2) y = 2x - 3
C) Shade the regions of solutions
1) The region that is over the line y = 4 - x
2) The region that is below the line y = 2x - 3
The solution is the intersection of both regions; this is the sector between both lines that is to the right of the intersection point, including the portion of the very line y = 4 - x and excluding the portion of the very line y = 2x - 3
Let U={1,2,3,4,5,6,7,8}A={2,3,6}B={2,4,5,7}C={1,7,8}Find the following use the roster method to write method to write the set. E
Ipatiy [6.2K]
A U B = set of all elements that are in both A or B:
A U B = {2, 3, 4, 5, 6, 7}
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Let s represent the length of any one side of the original square. The longer side of the resulting rectangle is s + 9 and the shorter side s - 2.
The area of this rectangle is (s+9)(s-2) = 60 in^2.
This is a quadratic equation and can be solved using various methods. Let's rewrite this equation in standard form: s^2 + 7s - 18 = 60, or:
s^2 + 7s - 78 = 0. This factors as follows: (s+13)(s-6)=0, so that s = -13 and s= 6. Discard s = -13, since the side length cannot be negative. Then s = 6, and the area of the original square was 36 in^2.
Answer:
A(2)
Step-by-step explanation:
