Local hardware store sells fencing for $1.30 per yard of each unit of a good represents 25 years how much will it cost to fix a problem presented by the polygon MNOP

(Please help me I really don’t understand )

Local hardware store sells fencing for 130 per yard of each unit of a good represents 25 years how much will it cost to fix a problem presented by the polygon M class=

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Answer:

Step-by-step explanation:

You go in order from MNOP, and you count the negatives and the positives and add them all together. I did that for you. But now add the negatives and positives, think of a number line, left of zero is negatives and right of zero is positive.(like I show in the second picture) make sense?

Ver imagen hanson50kellie
Ver imagen hanson50kellie

Answer:

The cost of fencing is $670.5.

Step-by-step explanation:

It is given that the cost of fencing is $1.30 per yard.

From the given figure it is clear that the M(-4,2), N(1,4), O(4,4), P(4,0).

Distance formula:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Using distance formula, we get

[tex]MN=\sqrt{\left(1-\left(-4\right)\right)^2+\left(4-2\right)^2}=\sqrt{29}[/tex]

[tex]NO=\sqrt{\left(4-1\right)^2+\left(4-4\right)^2}=3[/tex]

[tex]OP=\sqrt{\left(4-4\right)^2+\left(0-4\right)^2}=4[/tex]

[tex]MP=\sqrt{\left(4-\left(-4\right)\right)^2+\left(0-2\right)^2}=2\sqrt{17}[/tex]

The perimeter of the given polygon is

[tex]Perimeter=MN+NO+OP+MP[/tex]

[tex]Perimeter=\sqrt{29}+3+4+2\sqrt{17}[/tex]

[tex]Perimeter\approx 20.63[/tex]

The perimeter of the given polygon is 20.63 units.

It is given that 1 units = 25 yards.

[tex]20.63\text{ units}=20.63\times 25\text{ yards}[/tex]

[tex]20.63\text{ units}=515.75\text{ yards}[/tex]

Cost of fencing = $1.30 per yard

[tex]cost=670.475\approx 670.5[/tex]

Therefore the cost of fencing is $670.5.

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