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zubka84 [21]
3 years ago
14

A customer decides to build a fence around the backyard. The yard is rectangular, and with the house an equal distance from both

property lines. As a result, the fence will have five sections. The short sections, connecting to the house, are each 10 feet in length. The yard then has a depth of 65 feet with a width of 80 feet. How much fencing is needed?
Mathematics
1 answer:
DerKrebs [107]3 years ago
4 0

Answer:

D). 290 feet

Step-by-step explanation:

Length of the yard = 65 feet

Width of the yard = 80feet

The Perimeter of the yard = 2(l + b)

                                           = 2 * (65 + 80)

                                            = 2 * 145

                                             = 290 feet

∵ Fencing required = 290 feet

Thus, <u>option A</u> is the correct answer.

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1. y=5/4x
Then, make X the subject
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3. Once done, simply replace x with h(x)^-1 and y with x
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Sorry I'm not very good at rearranging formula but hope this helps :)
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What is the end behavior of the graph of the polynomial function f(x) = –x5 + 9x4 – 18x3?
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As x approaches ∞, y approaches -∞

As x approaches -∞, y approaches ∞

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f(x) = -x^5 + 9x^4 - 18x^3

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If exponent is odd and leading coefficient is negative then end behavior is

As x approaches ∞, y approaches -∞

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3 years ago
I need help with Q9 c, I’ve done a and b if it helps, b=325 but I am just stuck on question 9C, I know the answer is 500N (as it
777dan777 [17]

Step-by-step explanation:

I know you've already done parts a and b, but I'll show the work for that before I do c.

Draw two free body diagrams, one for the car and one for the trailer.  The car pulls the trailer forward with a tension force T, so the trailer pulls backward on the car with an equal and opposite force T.

The car also has a 1200 N forward force from the engine, and a 200 N backwards force from resistance.

The trailer has a backwards resistance force of 100 N.

Sum of forces on the car:

∑F = ma

1200 − 200 − T = 900a

1000 − T = 900a

Sum of forces on the trailer:

∑F = ma

T − 100 = 300a

To solve the system of equations, first add the equations together.

1000 − 100 = 1200a

900 = 1200a

a = 0.75 m/s²

Plug back into either equation to find the tension force:

T = 325 N

Now for part c, draw new free body diagrams for the car and trailer.  This time, the car is pushing back on the trailer to slow it down.  So the trailer is pushing forward on the car with an equal and opposite force.  The magnitude of that tension force is given to be 100 N.

The car also has a backwards 200 N force from resistance, and a backwards brake force F.

The trailer has a backwards 100 N force from resistance.

Sum of forces on the car:

∑F = ma

100 − 200 − F = 900a

-100 − F = 900a

Sum of forces on the trailer:

∑F = ma

-100 − 100 = 300a

-200 = 300a

a = -⅔

Plugging into the first equation:

-100 − F = 900 (-⅔)

-100 − F = -600

F = 500 N

4 0
2 years ago
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