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3 years ago

7

A rectangular flower bed that is 5 ft long and 4 feet wide is surrounded by a square lawn that is

1 answer:

mr Goodwill [35]3 years ago

6 0

Answer:

124 ft2

Step-by-step explanation:

The yard is a 12ft square so the length and width would both be 12ft. You multiply the length and width of the yard to get the area of the yard so 12x12=144. You have the length and width of the flower bed so 5x4=20. Once you have the area of both you can subtract the area the flower bed takes up of the lawn. 144-20=124 ft2.

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Solve for x. Round your answer to the nearest hundredth.<br><br>13.48x − 200 < 256.12

Alona [7]

Answer:

33.84

Step-by-step explanation:

Add 200 on both sides:

13.48x − 200 < 256.12

+200 +200

13.48x < 456.12

Divide by 13.48 on both sides:

13.48x < 456.12

/13.48 /13.48

x < 33.84

6 0

3 years ago

Find the perimeter of the polygon with the given vertices. Round your answer to the nearest hundredth.

Korolek [52]

Perimeter of the given polygon is 16.94 units.

To measure the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the ex[pression,

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

By using this expression,

Distance between A(0, 4) and B(2, 0) = $\sqrt{(0-2)^2+(4-0)^2}$

= $\sqrt{20}$ units

Distance between B(2, 0) and C(2, -2) = $\sqrt{(2-2)^2+(0+2)^2}$

= 2 units

Distance between C(2, -2) and D(0, -2) = $\sqrt{(0-2)^2+(-2+2)^2}$

= 2 units

Distance between D(0 -2) and E(-2, 2) = $\sqrt{(0+2)^2+(-2-2)^2}$

= $\sqrt{20}$ units

Distance between E(-2, 2) and F(-2, 4) = $\sqrt{(-2+2)^2+(2-4)^2}$

= 2 units

Distance between F(-2, 4) and A(0, 4) = $\sqrt{(-2-0)^2+(4-4)^2}$

= 2 units

Perimeter of ABCDEF = AB + BC + CD + DE + EF + FA

= $\sqrt{20}+2+2+\sqrt{20}+2+2$

= $8+2\sqrt{20}$

= 16.94 units

Therefore, perimeter of the given polygon in the graph attached is 16.94 units.

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2 years ago

Which points on the number line are 8 units away from 0?

IRINA_888 [86]

8 and -8 are the both 8 units away from 0 on a number line

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2 years ago

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How many cm of framing are needed to make a picture frame with outside measurements 40 cm and 25 cm

AnnZ [28]

Answer:

For average sized artwork (about 11x14" to 30x20"), the common mat width is 2". If you artwork is smaller, you might consider a 1.75 or 1.5" mat.

Step-by-step explanation:

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3 years ago

Solve for z.<br><br> x + y + z = 4<br><br> 4x - y - z = 1<br><br> x + z = 2

AleksAgata [21]

Answer:

z= 4-x-y

z= -1+4x-y

z= 2-x

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2 years ago

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