Question

What is the value of 4t2 + 6r - (t)(r) when t = -2 and r = 4?

Answer 1

$\huge\mathtt\colorbox{skyblue}{\color{purple}{AnSwEr}}$

$given \: t = - 2 \: \: and \: \: r = 4$

$= 4 {t}^{2} + 6r - (t)(r)$

$= 4 {( - 2)}^{2} + 6(4) - ( - 2)(4)$

$= 4(4) + 24 - ( - 8)$

$= 16 + 24 + 8$

${= }\bold\color{purple}{~48}$

Answer 2

$\huge{\textbf{\textsf{Question:}}}$

$\sf \color{red} \large What \: is \: the \: value \: of : \\ \bf \color{red} {4t}^{2} + 6r - (t)(r) \\ \sf \color{red} when \: \: \: \bf t = - 2 \: \: \: \sf and \: \: \: \bf r = 4$

$\huge{\textbf{\textsf{Answer:}}}$

$\bf \color{red} 4 {t}^{2} + 6r - (t)(r) \\ \bf \color{red} = 4( - 2)^{2} + 6(4) - ( - 2)(4) \\ \bf \color{red} = 4(4) + 24 - ( - 8) \\ \bf \color{red} = 16 + 24 + 8 \\ \bf \color{red} = \underline {\underline{48}}$

$\huge{\textbf{\textsf{Hope~it~helps..!!}}}$