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julsineya [31]
3 years ago
5

A dog is leashed to the corner of a house with an 18 foot leash. How much running area does the dog have?

Mathematics
2 answers:
navik [9.2K]3 years ago
8 0
If the dog were on a 15-ft leash tied to a stake in the ground 
out in the middle of a field, then he could cover every blade 
of grass in a 15-ft circle. The area of a circle is (pi R²), so 
he would have free range over (pi · 15²) square feet of area.

Sadly, he's not out in the middle of a field. He's tied to the corner 
of the house.  So he can't cover the whole circle, because the house 
blocks a quarter of the circle.  The dog only has access to

                       (3/4) of (pi R²)

                   =  (0.75) · π · (15²)

                   =  (0.75 · π · 225)  =  <span>530 square feet  </span>(rounded)
 
Natasha_Volkova [10]3 years ago
5 0
Find the area of 32 then you get it :)
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(7 + X) x 8 - 4 x X = 56
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Step-by-step explanation:

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Factor completely<br> x2+ 9x + 20
Sunny_sXe [5.5K]

Answer:

(x + 5) (x + 4)

Step-by-step explanation:

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x^2 + 9 x + 20

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4 0
3 years ago
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insens350 [35]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Pretty Pavers company is installing a driveway. Below is a diagram of the driveway they are
prohojiy [21]

Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

Therefore, the length of the sides of the square = √3 ft.

Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2

Where:

\theta = 2\times tan^{-1}( \frac{The \, number \, of  \, vertical  \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle}{The \, number \, of  \, horizontal \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle} )

\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214

Hence;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

4 0
3 years ago
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