Answer:
See below
Step-by-step explanation:
<h3>Part A</h3>
<h3>Part B</h3>
It has only one solution when the line mx + 1 is parallel to one part of |2x - 4|.
As per graph, we see m = 2 makes it parallel to y = 2x - 4. This is the only solution.
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Now, as I understand, you are looking for 2 solutions instead of one. It means the line mx + 1 will have 2 points of intersection with the line |2x - 4|.
One of those solutions is attached as well, where m = 1. And you may get many more options for this.
Answer:
x=3
Step-by-step explanation:
1/4x=1/2+1/4
1/4x=3/4
4*1/4x=3/4*4
x=3
SOLUTION:
A box has a square base of side x and height y where x,y > 0.
Its volume is V = x^2y and its surface area is
S = 2x^2 + 4xy.
(a) If V = x^2y = 12, then y = 12=x^2 and S(x) = 2x^2 C 4x (12=x2) = 2x2 + 48x^-1. Solve S'(x) = 4x - 48x-2 = 0 to
obtain x = 12^1/3. Since S(x/) ---> infinite as x ---> 0+ and as x --->infinite, the minimum surface area is S(12^1/3) = 6 (12)^2/3 = 31.45,
when x = 12^1/3 and y = 12^1/3.
(b) If S = 2x2 + 4xy = 20, then y = 5^x-1 - 1/2 x and V (x) = x^2y = 5x - 1/2x^3. Note that x must lie on the closed interval [0, square root of 10]. Solve V' (x) = 5 - 3/2 x^2 for x>0 to obtain x = square root of 30 over 3 . Since V(0) = V (square root 10) = 0 and V(square root 30 over 3) = 10 square root 30 over 9 , the
maximum volume is V (square root 30 over 3) = 10/9 square root 30 = 6.086, when x = square root 30 over 3 and y = square root 30 over 3 .
5. The graph of g(x) is narrower. Both graphs open upward. The vertex of g(x), (0,10), is translated 10 units up from the vertex of f(x) at (0,0)
6. The graph of g(x) is wider. Both graphs open upward. The vertex of g(x), (0,-3) is translated 3 units down from the vertex of f(x) at (0,0)
7.The graph of g(x) is narrower. g(x) opens downward and f(x) opens upward. The vertex of g(x), (0,8) is translated 8 units up from the vertex of f(x) at (0,0).
8. The graph of g(x) is wider. g(x) opens downward and f(x) opens upward. The vertex of g(x), (0,1/4) is translated 1/4 units up from the vertex of f(x) at (0,0).
9. A. h1(t)=-16t^2+400 h2(t)= -16t^2+1600
9. B The graph of h2 is a vertical translation of the graph of h1 : 1200 units up.
9. C sandbag dropped from 400 ft: 5 s
sandbag dropped from 1600 ft: 10 s
