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2 years ago

If a fence has posts 8 feet apart, how many posts are required to build a 40 foot long fence?

Mathematics

1 answer:

mamaluj [8]2 years ago

3 0

there would have to be 5 posts

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Can you tell me which one plz ?

Andreas93 [3]

Just saying you have a 25% chance of getting it right if you guess

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2 years ago

PLEASE HELP!!!! Show steps and explain how to get the answer please <br><br> Solve 5x-16=4(x-8)-3x

Anvisha [2.4K]

5x-16 = 4(x-8)-3x

first expand what is in parenthesis

4(x-8) = 4x-32

so now you have 5x-16 = 4x-32-3x

subtract 3x from 4x on the right side of equation 4x-3x = x

5x-16 = x -32

subtract 1x from each side to get

4x-16 = -32

add 16 to both sides to get 4x=-16

x = -16/4 = -4

x=-4

6 0

2 years ago

Read 2 more answers

Help with 21 would be much appreciated.

umka2103 [35]

Answer:

$\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}$

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

$\dfrac{AD}{CD}=\dfrac{CD}{DB}$

We have

$AD=14-8=6\\CD=y\\DB=8$

Substitute:

$\dfrac{6}{y}=\dfrac{y}{8}$ <em>cross multiply</em>

$y^2=(6)(8)$

$y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}$

For x use the Pythagorean theorem:

$x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}$

3 0

2 years ago

I dont understand this can someone give me the answer. "Determine the equation of the line represented on the graph below.

erma4kov [3.2K]

Answer:

y=2/5x-1

Step-by-step explanation:

7 0

2 years ago

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Steve and Abby purchased a set of vases to place on a 12 foot long mantel above their fireplace. They want to place one vase 1/4

Nonamiya [84]

Answer: One vase should be placed 3 feet from the end of the mantel and the other vase should be placed 9 feet from the same end.

Step-by-step explanation:

You know that lenght of the mantel above their fireplace is 12 feet.

They want to place one vase $\frac{1}{4}$ of the distance from one end of the mantel. This distance in feet is:

$distance_{vase1}=(12feet)(\frac{1}{4})\\\\distance_{vase1}=3feet$

They want to place the other vase $\frac{3}{4}$ of the distance from the same end of the mantel. Therefore, this distance in feet is:

$distance_{vase2}=(12feet)(\frac{3}{4})\\\\distance_{vase2}=\frac{36feet}{4}\\\\distance_{vase2}=9feet$

Therefore, one vase should be placed 3 feet from the end of the mantel and the other vase should be placed 9 feet from the same end.

5 0

2 years ago