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i inserted a picture of the question, i can give you the answer to the previous question if it helps. C(t) = -0.30(t-12)^2 + 40

I Inserted A Picture Of The Question I Can Give You The Answer To The Previous Question If It Helps Ct 030t122 40 class=

Sagot :

Answer:

F(t) = -0.54(t - 12)² + 104

Explanation:

We know that C(t) = -0.30(t - 12)² + 40 and F(t) = 9/5C(t) + 32

Then, we can replace C(t) on the equation of F(t) to get

[tex]\begin{gathered} F(t)=\frac{9}{5}C(t)+32 \\ F(t)=\frac{9}{5}(-0.30(t-12)^2+40)+32 \\ F(t)=\frac{9}{5}(-0.30)(t-12)^2+\frac{9}{5}(40)+32 \\ F(t)=-0.54(t-12)^2+72+32 \\ F(t)=-0.54(t-12)^2+104 \end{gathered}[/tex]

Therefore, the new function F(t) is

F(t) = -0.54(t - 12)² + 104

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