What are 3 different equations that have x=5 as a solution
2 answers:
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Part.1
Given equation:
C(t) = -0.30 (t – 12)²<span> + 40
</span>For t = 0
C(t) = -0.30 (0 - 12)² + 40
C(t) = -0.30 (-12)² + 40
C(t) = -3.2<span>
</span><span>For t = 12 (noon)
C(t) = -0.30 (12 - 12)</span>² + 40
C(t) = -0.30 (0)² + 40
C(t) = 40
<span>For t = 24 (midnight)
</span>C(t) = -0.30 (24 - 12)² + 40
C(t) = -0.30 (12)² + 40
C(t) = -0.30 × 144 + 40
C(t) = - 43.2 + 40
C(t) = -3.2
Part.2
Graph of the equation is given in the attachment.
Part.3
C(t) = –0.30(t – 12)²<span> + 40
F(t)=9/5C(t)+32
Substituting the values:
</span>F(t)=9/5{–0.30 (t – 12)² + 40}+32
F(t) = -0.54 (t – 12)² + 72 + 32
F(t) = -0.54 (t – 12)² + 104
Given equation:
C(t) = -0.30 (t – 12)²<span> + 40
</span>For t = 0
C(t) = -0.30 (0 - 12)² + 40
C(t) = -0.30 (-12)² + 40
C(t) = -3.2<span>
</span><span>For t = 12 (noon)
C(t) = -0.30 (12 - 12)</span>² + 40
C(t) = -0.30 (0)² + 40
C(t) = 40
<span>For t = 24 (midnight)
</span>C(t) = -0.30 (24 - 12)² + 40
C(t) = -0.30 (12)² + 40
C(t) = -0.30 × 144 + 40
C(t) = - 43.2 + 40
C(t) = -3.2
Part.2
Graph of the equation is given in the attachment.
Part.3
C(t) = –0.30(t – 12)²<span> + 40
F(t)=9/5C(t)+32
Substituting the values:
</span>F(t)=9/5{–0.30 (t – 12)² + 40}+32
F(t) = -0.54 (t – 12)² + 72 + 32
F(t) = -0.54 (t – 12)² + 104