Please help me ASAP!!!

asked Feb 25, 2023 128k views

3 votes

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4 votes

$f(x)=\sqrt[]{2x^2-3x+\text{ 1}}$

substitute x = 5 in the above function

$f(5)=\sqrt[]{2(5)^2-3(5)+1}$
$=\sqrt[]{2(25)-15+1}$
$=\sqrt[]{50-15+1}$
$=\sqrt[]{36}=\text{ 6}$

f(5) = 6

Tomba answered Mar 4, 2023

by Tomba

3.2k points

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