Ask question

Ask question

All categories

3 years ago

15

A square and a rectangle have the same perimeters. The length of a side of the square is 4x-1. The length of the rectangle is 2x

+1 and the width is x+2. Write and solve an equation to find x.

1 answer:

SIZIF [17.4K]3 years ago

6 0

Perimeter of Rectangle= Perimeter of square

2(L+B) = 4*L
2(2x+1 + x+2) = 4*(4x-1)
6x+6= 16x -4
10x = 10
x = 1

You might be interested in

Please Help Me as soon as possible

disa [49]

Answer:

B bc 3 would be in between x and 5, and x equals 0

Step-by-step explanation:

hopefully this helped! :)

3 0

2 years ago

Read 2 more answers

Please help quickly !!!

kenny6666 [7]

Answer:

B) w^35

Step-by-step explanation:

you multiple the power of 5 by 7 to get 35

8 0

2 years ago

Read 2 more answers

lakkis [162]

where is your triangle mate??

hope this helps u

follow me

mark as brainliest

thank me

5 0

3 years ago

SIZIF [17.4K]

Answer:

bcd= congruent property abstracted by -cdb

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is 1 6/9 +5 9/7

lisabon 2012 [21]

$\longrightarrow{\green{7 \frac{20}{21}}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

$1 \frac{6}{9} + 5 \frac{9}{7} \\ \\ = \frac{15}{9} + \frac{44}{7} \\ \\ = \frac{15 \times 7}{9 \times 7} + \frac{44 \times 9}{7 \times 9} \\ \\ = \frac{105 + 396}{63} \\ \\ = \frac{501}{63} \\ \\ = \frac{167}{21} \\ \\ = 7 \frac{20}{21}$

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{♡}}}}}$

6 0

3 years ago