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erastova [34]
3 years ago
8

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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PLEASE HELP ME !! WILL GIVE BRAINLIEST!!
PolarNik [594]
B is a times b
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6 0
3 years ago
Answer it please and thank you..
Kruka [31]

(-6,2) Is the answer! :)


4 0
3 years ago
the length of a rectangle is 9 centimeters more than the width. the area is 112 square centimeters. find the length and width of
NeX [460]
Here,
let width(b)be x then,
length (l)=9cm+x
area =112 sq cm
now,
area of rectangle=l*b
or, 112=(9+x)x
or, 112=9x+x^2
or, 0=x^2+9x-112
or, 0=x^2+(16-7)x-112
or, 0=x^2+16x-7x-112
or, 0=x(x+16)-7(x+16)
or, 0=(x-7)(x+16)
either,
0=x-7
or,7=x
x=7cm
Or,
0=x+16
or, -16=x
x= -16[impossible,as distance is never negative] so,
x=7cm
therefore,length = 7cm + 9 cm = 16cm and width = 7cm.
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8 0
3 years ago
Read 2 more answers
Is it a function and what is the Domain and the Range
Schach [20]

Answer:

a-is a function

the domain is:[-3,1]

the range is :[-4,0]

b- is not a function

7 0
3 years ago
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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}
6 0
3 years ago
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