9. a rectangle has a width that is 2 feet longer than four times the length. what is the length of the rectangle
if the perimeter is 64 feet? (hint: draw a picture and use algebra to solve)

6 feet

Step-by-step explanation:

Perimeter = 2L + 2W

Since you know the perimeter is 64 feet, you would plug that into the perimeter side.

64 = 2L + 2W

You also know the width is 2 feet longer than 4 times the length.

64 = 2L + 2(2 + 4L)  > Multiply 2 * 2 and 2 * 4l

64 = 2L + 4 + 8L  > Subtract 4 from each side

-4           -4

60 = 2L + 8l > add these two

60 = 10L

/10   /10 > divide by 10 on each side

6 = L

Now that we know the length, find the width.

64 = 2(6) + 2(2 + 4(6)) > multiply 2 * 6 and 4 * 6

64 = 12 + 2(2 + 24) > add 24 and 2

64 = 12 + 2(26) > multiply 2 and 26

Before you go any further, you now know that the width is 26 feet. Continue solving.

64 = 12 + 52

64 = 64

Everything checks out.

The length is 6 feet and the width is 26 feet.

this is answer of 144 cubic meters

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15