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What is the value of ๐‘ฆy when ๐‘ฆ=3๐‘ฅ2+2๐‘ฅโˆ’5y=3x 2 +2xโˆ’5 and ๐‘ฅ=โˆ’1x=โˆ’1?

Question

What is the value of ๐‘ฆy when ๐‘ฆ=3๐‘ฅ2+2๐‘ฅโˆ’5y=3x 2 +2xโˆ’5 and ๐‘ฅ=โˆ’1x=โˆ’1?

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Solution

To find the value of y, we need to substitute x = -1 into the equation y = 3x^2 + 2x - 5.

Step 1: Substitute x = -1 into the equation y = 3(-1)^2 + 2(-1) - 5

Step 2: Simplify the equation y = 3(1) - 2 - 5 y = 3 - 2 - 5

Step 3: Solve the equation y = -4

Similar Questions

To find the ๐‘ฆy value of the equation ๐‘ฆ=5๐‘ฅy=5x when ๐‘ฅ=โˆ’1x=โˆ’1:Given the equation:๐‘ฆ=5๐‘ฅy=5xSubstitute ๐‘ฅ=โˆ’1x=โˆ’1 into the equation:๐‘ฆ=5(โˆ’1)y=5(โˆ’1)๐‘ฆ=โˆ’5y=โˆ’5So, the ๐‘ฆy value when ๐‘ฅ=โˆ’1x=โˆ’1 is โˆ’5โˆ’5.

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When ๐‘“(๐‘ฅ) = 3๐‘ฅ โˆ’ 5 and ๐‘”(๐‘ฅ) = 2๐‘ฅ2 โˆ’ 5 , find ๐‘“(๐‘”(๐‘ฅ))

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