The length of a rectangle is 1 foot more than twice the width the perimeter is 44 feet find the dimensions of the rectangle

The length of rectangle is 15 feet and width of rectangle is 7 feet

Solution:

Let "a" be the width of rectangle

Given that, length of rectangle is 1 foot more than twice the width

Therefore, we get

length = 1 + twice the width

length = 1 + 2a eqn 1

Given that perimeter is 44 feet

The perimeter of rectangle is given as:

Perimeter = 2(length + width)

44 = 2(1 + 2a + a)

44 = 2(1+3a)

22 = 1 + 3a

3a = 22 - 1

3a = 21

a = 7

Thus from eqn 1,

length = 1 + 2(7) = 1 + 14 = 15

Thus length of rectangle is 15 feet and width of rectangle is 7 feet

answer: the first one

step-by-step explanation: use a calculator and divide the minutes by the dollars.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3