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Maurinko [17]
2 years ago
14

Which number is a common multiple of 2 and 4?

Mathematics
1 answer:
katovenus [111]2 years ago
4 0

Answer:

Step-by-step explanation:

c I think I did the math that’s what it leads to so yeah

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the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 58 in, find its length and width
dolphi86 [110]

Answer:

  • Length = 17 inches

  • Width = 12 inches

⠀

Step-by-step explanation:

⠀

As it is given that, the length of a rectangle is 5 in longer than its width and the perimeter of the rectangle is 58 in and we are to find the length and width of the rectangle. So,

⠀

Let us assume the width of the rectangle as x inches and therefore, the length will be (x + 5) inches .

⠀

Now, <u>According to the Question :</u>

⠀

{\longrightarrow \qquad { \pmb{\frak {2 ( Length + Breadth )= Perimeter_{(Rectangle)} }}}}

⠀

{\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{2 ( 2x + 5  )= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x = 58  - 10}}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x = 48}}}}

⠀

{\longrightarrow \qquad { {\sf{ x =  \dfrac{48}{4} }}}}

⠀

{\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\:  \:  \bigstar

⠀

Therefore,

  • The width of the rectangle is 12 inches .

⠀

Now, We are to find the length of the rectangle:

{\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}

⠀

{\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}

⠀

{\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}

⠀

Therefore,

  • The length of the rectangle is 17 inches .

⠀

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