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3 years ago

The rectangle below has an area of 30k3 + 6k2.

Mathematics

1 answer:

Anit [1.1K]3 years ago

8 0

9514 1404 393

Answer:

length: (5k +1)
width: 6k^2

Step-by-step explanation:

The GCF of the two terms is the second term: 6k^2. When that is factored out, you have ...

area = (6k^2)(5k +1)

According to the problem statement, this is interpreted as ...

width: 6k^2

length: 5k+1

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the length of a rectangle is 9 centimeters more than the width. the area is 112 square centimeters. find the length and width of

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Here,
let width(b)be x then,
length (l)=9cm+x
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area of rectangle=l*b
or, 112=(9+x)x
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The width of a rectangle is 6 2/3 inches. The length of the is twice it’s width. What so the perimeter of the rectangle?

Nookie1986 [14]

$\text {Width = } 6\dfrac{2}{3} \text { inches}
$

The length is twice its width:
$\text {Length = } 2 \times 6\dfrac{2}{3} \text { inches}$

Change to improper fraction:
$\text {Length = }2 \times \dfrac{20}{3} \text { inches}$

Combine into single fraction:
$\text {Length = } \dfrac{40}{3} \text { inches}$

Find Perimeter :
$\text {Perimeter = Length + Length + Width + Width}$

$\text {Perimeter = } \dfrac{40}{3} + \dfrac{40}{3} + \dfrac{20}{3} + \dfrac{20}{3} = \dfrac{120}{3} = 40 \text { inches}$

$\bf \text {Answer: Perimeter = 40 inches}$

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