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3 years ago

The length and width of a rectangle are ( 2x -5) and ( 3 x +4) What is the Perimeter?

Mathematics

1 answer:

Stella [2.4K]3 years ago

5 0

Answer:

$\fbox{ \sf{Perimeter \: of \: a \: rectangle = 10x - 2}}$

Step-by-step explanation:

$\star{ \sf{ \: \: Length \: of \: a \: rectangle ( l ) = (2x - 5) }}$

$\star{ \sf{ \: Width \: of \: a \: rectangle ( w ) = (3x + 4)}}$

$\star{ \sf{ \: \: Perimeter \: of \: a \: rectangle = ?}}$

$\boxed{ \bold{ \sf{Perimeter \: of \: a \: rectangle = 2(l + w)}}}$

$\hookrightarrow{ \sf{Perimeter \: of \: a \: rectangle = 2(2x - 5 + 3x + 4)}}$

$\text{Step \: 1 : \: Collect \: like \: term}$

$\text{Like \: terms \: are \: those \: which \: have \: the \: same \: base}$

$\hookrightarrow{ \sf{2(2x + 3x - 5 + 4)}}$

$\hookrightarrow{ \sf{2(5x - 5 + 4)}}$

$\text{Step \: 2 \: : \: Subtract \: 4 \: from \: 5}$

$\text{Remember !\: : the \: negative \: and \: positive \: integers \: are \: always \: subtracted \: but \: posses \: the \: sign \: of \: the \: bigger \: integer}$

$\hookrightarrow{ \sf{2(5x - 1)}}$

$\text{Step \: 3 \: : \: Distribute \: 2 \: through \: the \: parentheses}$

$\hookrightarrow{ \sf{2 \times 5x - 2 \times 1}}$

$\hookrightarrow{ \sf{10x - 2}}$

$\sf{∴ \: The \: perimeter \: of \: a \: rectangle = 10x - 2}$

$\text{Hope \: I \: helped !}$

$\text{Best \: regards ! }$

~ $\text{TheAnimeGirl}$

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3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E to the system of inequalities from part A, and check if its is true, what we get.

$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to our system of inequalities.

Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).