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nexus9112 [7]
3 years ago
9

What is the range of the function y=3^x x>0 R y<0 y>0

Mathematics
1 answer:
Dima020 [189]3 years ago
5 0

Answer:

The range is  y > 0

Step-by-step explanation:

3^x is always greater than 0 and cannot be negative   unless x is not real

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<h2>Explanation:</h2>

Given the equation expressed as:

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Step 1: Given the equation 2x + 5 = 8x

Step 2: Subtract 5 from both sides:

\begin{gathered} 2x+5-5=8x-5 \\ 2x+\cancel{5}-\cancel{5}=8x-5 \\ 2x=8x-5 \end{gathered}

Step 3: Subtract 8x from both sides

\begin{gathered} 2x-8x=8x-8x-5 \\ -6x=\cancel{8x}-\cancel{8x}-5 \\ -6x=-5 \end{gathered}

Step 4: Divide both sides by -6

\begin{gathered} \frac{-6x}{-6}=\frac{-5}{-6} \\ x=\frac{5}{6} \end{gathered}

Step 5: Get the value of 12x. Substitute x = 5/6 into the expression to have:

\begin{gathered} 12x=12(\frac{5}{6}) \\ 12x=\cancel{12}^2(\frac{5}{\cancel{6}^1}) \\ 12x=2\times5 \\ 12x=10 \end{gathered}

Therefore the value of 12x is 10

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