Evaluate the expression using the given values.
StartAbsoluteValue 6 x squared minus 3 y EndAbsoluteValue
;
xequals
5,
yequals
3
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Part 1
StartAbsoluteValue 6 x squared minus 3 y EndAbsoluteValue
equalsenter your response here
Let's first substitute the given values of x and y into the expression.
Given that \( x = 5 \) and \( y = 3 \), our expression becomes:
\[ \lvert 6x^2 - 3y \rvert \]
Now we substitute \( x = 5 \) and \( y = 3 \):
\[ \lvert 6(5)^2 - 3(3) \rvert \]
\[ \lvert 6(25) - 9 \rvert \]
\[ \lvert 150 - 9 \rvert \]
\[ \lvert 141 \rvert \]
The absolute value of 141 is just 141, since it's already positive:
\[ \lvert 141 \rvert = 141 \]
Thus, the answer is:
\[ \lvert 6x^2 - 3y \rvert = 141 \]