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Neporo4naja [7]
3 years ago
5

HEYYY PLZ HELP ME BESTIES TYSM ILY

Mathematics
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

<em>y-4=-2(x+11)</em>

Step-by-step explanation:

First substitute:

Y-y1=m(x-x1)

y1= the y coordinate of a point

m= slope

x1= the x coordinate of a point

Using that, substitute:

y-4=-2(x--11)

y-4=-2(x+11)

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Answer:

Given that

sinθ = 0.8660

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⇒ θ = sin⁻¹(√3/2)

⇒ θ = 60°

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Find the circumference of a circle if the radius is 15 ft. Use 3.14 for pi.
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Circumference = 2 x radius x Pi

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Circumference = 94.2 feet

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I'm giving 50 points for this ​
olchik [2.2K]

Answer:

STAN LOONA

Step-by-step explanation:

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3 years ago
A study of injuries to in-line skaters used data from the National Electronic Injury Surveillance System, which collects data fr
Katarina [22]

Answer:

(a)  One count is only 7, and the guidelines for using the large-sample method call for all counts to be at least 10.

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Attached is the solution to b and c

8 0
3 years ago
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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