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siniylev [52]
3 years ago
9

You can graph any 2d linear function if you know the slope and the y-intercept. True False

Mathematics
1 answer:
Dmitry [639]3 years ago
3 0
True and it’s the best answer hope it helps.
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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.

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Regular dodecahedron with s= 2 times the square root of three and a = 2.4
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Here are two steps from the derivation of the quadratic formula,
BaLLatris [955]

Answer:

A

Step-by-step explanation:

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The distance between any point (x0,y0) on the parabola and the focus (m,n) is the same as the distance between (x0,y0) and the directrix line ax+by+c. The distance between (x0,y0) and focus (a,b) is \sqrt((x-m)^2+(y-n)^2). The distance between (x0,y0) and ax+by+c is |ax0+by0+c|/\sqrt(m^2+n^2).  Equalize these two expressions.
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105 divided by 3<br> long division estimating please show your work
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35 I had this for a test the answer is

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