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svetlana [45]
3 years ago
12

What values of b satisfy 4(3b + 2)2 = 64?

Mathematics
1 answer:
Vikentia [17]3 years ago
8 0
Given the equation 4(3b + 2)² = 64, dividing both sides of the equation by 4, we have (3b + 2)² = 16 and getting the square root of both sides, (3b + 2) = 4 and (3b + 2) = -4 We can solve for b for each equation and have 3b = 2 | 3b = -6 b = 2/3 | b = -2 Therefore, the values of b are 2/3 and -2 and from the choices, the answer is <span>A: b = 2/3 and b = -2.</span>
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To find the minimum value, graph each of the inequalities. After graphing each inequality, test a point and shade the region that satisfies the inequality. Once all inequalities have been shaded, find the region where they all overlap. The region will be bounded by intersection points. Test each of these points into C=x+3y. The least value for C is the minimum.

(14,0)                  (0,17.5)                   (3.08,3.64)

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