Someone help me as fast as possible!!

Question

asked Dec 5, 2023 80.6k views

1 Answer

Dylan Harness · Answer 1 · 2023-12-10T03:11:20+0000

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: length = 11 \: \: in$

$\qquad \tt \rightarrow \: width = 4 \: \: in$

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$\large \tt Solution \: :$

Let the measures be :

$\textsf{\large Area of rectangle :}$

$\qquad \tt \rightarrow \: Area = length \sdot width$

$\qquad \tt \rightarrow \: 44 = x \sdot(x + 7)$

$\qquad \tt \rightarrow \: {x}^(2) + 7x = 44$

$\qquad \tt \rightarrow \: {x}^(2) + 7x - 44 = 0$

$\qquad \tt \rightarrow \: {x}^(2) + 11x - 4x - 44 = 0$

$\qquad \tt \rightarrow \: x(x + 11) - 4(x + 11) = 0$

$\qquad \tt \rightarrow \: (x + 11)(x - 4) = 0$

possible values of x = -11 and 4

$\textsf{But the acceptable value of x = 4}$

[ length or width of rectangle can't be negative ]

$\qquad \tt \rightarrow \: length = x + 7 = 4 + 7 = 11 \: \: in$

$\qquad \tt \rightarrow \: width = x = 4 \: \: in$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞