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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

Question

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

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Solution

It seems like there's a bit of confusion in your question as the function g(t) is defined twice with different expressions. However, if we're looking for the value of g(t) when t=3, we should use the second definition.

The second definition of g(t) is: g(t) = t - 3/2t^2 + 2t - 24

Let's substitute t=3 into this equation:

g(3) = 3 - 3/2*(3^2) + 23 - 24 = 3 - 3/29 + 6 - 24 = 3 - 13.5 + 6 - 24 = -28.5

So, the answer is not listed among the options you provided (A.14, B.3, C.7, D.None of these, E.0). The correct answer should be -28.5. Please check the question again.

This problem has been solved

Similar Questions

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If the following function is continuous, then what is the value of b?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

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

b. 3t2ย +ย 24tย +ย 45ย =ย 0

(b) ๐‘“ (๐‘ฅ + ๐‘–๐‘ฆ) = ๐‘ฅ3 + ๐‘–(๐‘ฆ โˆ’ 1)3

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