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A.๐‘“(๐‘ก)=2๐‘กโˆ’๐‘ก3f(t)=2 t โˆ’t 3 B.๐‘‘(๐‘ก)=(1.1)๐‘กd(t)=(1.1) t C.๐‘(๐‘ก)=๐‘ก4โˆ’3๐‘ก+9b(t)=t 4 โˆ’3t+9D.โ„Ž(๐‘ก)=5๐‘ก+๐‘ก5h(t)=5 t +t 5 E.๐‘(๐‘ก)=๐‘ก2โˆ’5๐‘กc(t)= t 2 โˆ’5tโ€‹

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Solution 1

To analyze the functions provided, I will break down each function into smaller parts, apply relevant concepts, analyze the functions in detail, and provide a concise summary of findings.

1. Break Down the Problem

The functions are polynomial expressions defined with respect to t t :

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Similar Questions

A.๐‘“(๐‘ก)=2๐‘กโˆ’๐‘ก3f(t)=2 t โˆ’t 3 B.๐‘‘(๐‘ก)=(1.1)๐‘กd(t)=(1.1) t C.๐‘(๐‘ก)=๐‘ก4โˆ’3๐‘ก+9b(t)=t 4 โˆ’3t+9D.โ„Ž(๐‘ก)=5๐‘ก+๐‘ก5h(t)=5 t +t 5 E.๐‘(๐‘ก)=๐‘ก2โˆ’5๐‘กc(t)= t 2 โˆ’5tโ€‹

๐‘”(๐‘ก)={2๐‘ก2+2๐‘กโˆ’24๐‘กโˆ’3๐‘–๐‘“ย ๐‘กโ‰ 3๐‘๐‘–๐‘“ย ๐‘ก=3g(t)={ tโˆ’32t 2 +2tโˆ’24โ€‹ bโ€‹ ifย tโ‰ 3ifย t=3โ€‹ A.14B.3C.7D.None of theseE.0

Which function defines (๐‘“รท๐‘”)โข(๐‘ฅ) ?๐‘“โก(๐‘ฅ)=(3.6)๐‘ฅ+2๐‘”โก(๐‘ฅ)=(3.6)3โข๐‘ฅ+1 A. (๐‘“รท๐‘”)โข(๐‘ฅ)=(3.6)-2โข๐‘ฅ+1 B. (๐‘“รท๐‘”)โข(๐‘ฅ)=(3.6)4โข๐‘ฅ+3 C. (๐‘“รท๐‘”)โข(๐‘ฅ)=(1.8)3โข๐‘ฅ2+7โข๐‘ฅ+2 D.

For the given A.P. a = 3.5, d = 0, then tn = ยทยทยทยทยทยทยทยทยทยทยทยทยทยท ยท(A) 0/gw 3.5(C) 103.5(D) 104.5

The range of ๐‘“(๐‘ฅ)=โˆฃ๐‘ฅโˆ’3โˆฃf(x)=โˆฃxโˆ’3โˆฃis:A.None of theseB.๐‘ฆโ‰ฅ3yโ‰ฅ3C.๐‘ฆ<3y<3D.๐‘ฆ>3y>3E.๐‘ฆโ‰ฅ0yโ‰ฅ0

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