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lianna [129]
3 years ago
14

X/2 - y/3 = 3/2X/3 + y/2 = 16/3

Mathematics
1 answer:
Ivenika [448]3 years ago
4 0
X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)

x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6
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To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq  0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq  0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible  \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)

Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq  0\\-112 \geq  0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq  0\\80 \geq  0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq  0\\-112 \geq  0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution

Conclusion:

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