SosieT692633
SosieT692633
25-10-2022
Mathematics
contestada
please help me ASAP!!!
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KemoniO421682
KemoniO421682
25-10-2022
[tex]f(x)=\sqrt[]{2x^2-3x+\text{ 1}}[/tex]
substitute x = 5 in the above function
[tex]f(5)=\sqrt[]{2(5)^2-3(5)+1}[/tex][tex]=\sqrt[]{2(25)-15+1}[/tex][tex]=\sqrt[]{50-15+1}[/tex][tex]=\sqrt[]{36}=\text{ 6}[/tex]
f(5) = 6
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