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Using the properties of equality, find the solution to the equation 2f5−85+2f=8 .(1 point) Responses 4 4 83 Start Fraction 8 ov...Question
                Using the properties of equality, find the solution to the equation 2f5−85+2f=8.(1 point)
Responses
83
Start Fraction 8 over 3 End Fraction
12
12
4
4
166
            
        Responses
83
Start Fraction 8 over 3 End Fraction
12
12
4
4
166
Answers
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                    Answered by
            GPT-4o mini
            
    To solve the equation \(2f^5 - 85 + 2f = 8\), we can start by rearranging the equation. 
1. First, let's isolate the terms that include \(f\):
\[
2f^5 + 2f - 85 = 8
\]
2. Next, we move \(8\) to the left side of the equation:
\[
2f^5 + 2f - 85 - 8 = 0
\]
This simplifies to:
\[
2f^5 + 2f - 93 = 0
\]
3. Now you can factor out the common term \(2\) in the equation:
\[
f^5 + f - 46.5 = 0
\]
Since this is a polynomial equation, we can try to find rational solutions or use numerical methods or graphing techniques for approximations.
We will test the provided options to see which might be a solution.
- Checking \(f = 4\):
\[
f^5 + f - 46.5 \rightarrow 4^5 + 4 - 46.5 = 1024 + 4 - 46.5 = 981.5 \quad (\text{not } 0)
\]
- Checking \(f = 12\):
\[
12^5 + 12 - 46.5 = 248832 + 12 - 46.5 = 248797.5 \quad (\text{not } 0)
\]
- Checking \(f = \frac{8}{3}\):
\[
\left(\frac{8}{3}\right)^5 + \frac{8}{3} - 46.5 \rightarrow \frac{32768}{243} + \frac{8}{3} - 46.5
\]
This computation is more complex; thus we will do the easier checks first.
- Checking \(83\):
\[
83^5 + 83 - 46.5 \quad \text{(too big, unlikely 0)}
\]
- \(166\):
\[
166^5 + 166 - 46.5 \quad \text{(too big, unlikely 0)}
\]
Based on tests, we see right away that none of these numbers likely come up to 0. If we eventually couldn't verify simple plug-ins easily, we would need to use synthetic division or numerical methods to find a root since these seem much higher than simple plug-ins.
However, examining logical ranges, given the cubic nature of the problem, test potential simplifications. Using simple trial, numerical substitutions, calculus or graphing may narrow values, but at odds hence need numerical estimates away from mere plug tests.
Given that you provided responses, it'd suggest further validating to hint that solutions weren't elegantly rounding inside those variables. You might check for miswritten equalities or estimation at calculus graphs.
    
1. First, let's isolate the terms that include \(f\):
\[
2f^5 + 2f - 85 = 8
\]
2. Next, we move \(8\) to the left side of the equation:
\[
2f^5 + 2f - 85 - 8 = 0
\]
This simplifies to:
\[
2f^5 + 2f - 93 = 0
\]
3. Now you can factor out the common term \(2\) in the equation:
\[
f^5 + f - 46.5 = 0
\]
Since this is a polynomial equation, we can try to find rational solutions or use numerical methods or graphing techniques for approximations.
We will test the provided options to see which might be a solution.
- Checking \(f = 4\):
\[
f^5 + f - 46.5 \rightarrow 4^5 + 4 - 46.5 = 1024 + 4 - 46.5 = 981.5 \quad (\text{not } 0)
\]
- Checking \(f = 12\):
\[
12^5 + 12 - 46.5 = 248832 + 12 - 46.5 = 248797.5 \quad (\text{not } 0)
\]
- Checking \(f = \frac{8}{3}\):
\[
\left(\frac{8}{3}\right)^5 + \frac{8}{3} - 46.5 \rightarrow \frac{32768}{243} + \frac{8}{3} - 46.5
\]
This computation is more complex; thus we will do the easier checks first.
- Checking \(83\):
\[
83^5 + 83 - 46.5 \quad \text{(too big, unlikely 0)}
\]
- \(166\):
\[
166^5 + 166 - 46.5 \quad \text{(too big, unlikely 0)}
\]
Based on tests, we see right away that none of these numbers likely come up to 0. If we eventually couldn't verify simple plug-ins easily, we would need to use synthetic division or numerical methods to find a root since these seem much higher than simple plug-ins.
However, examining logical ranges, given the cubic nature of the problem, test potential simplifications. Using simple trial, numerical substitutions, calculus or graphing may narrow values, but at odds hence need numerical estimates away from mere plug tests.
Given that you provided responses, it'd suggest further validating to hint that solutions weren't elegantly rounding inside those variables. You might check for miswritten equalities or estimation at calculus graphs.
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