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Brums [2.3K]
3 years ago
13

HELPPPPPPPPP ASAPPPPPPPP PLSSSS

Mathematics
1 answer:
yanalaym [24]3 years ago
8 0

-1/3(px-2) = 9

ok! so first we want to isolate x.

multiply both sides by -3/1 (this is the reciprocal of -1/3)

(px-2) = -27

px = -27 + 2

px = -25

x = -25/p

hope this helped!

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abruzzese [7]

\huge{\boxed{2 \sqrt{101}}}

The distance formula is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, where (x_1, y_1) and (x_2, y_2) are the points.

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Solve the exponents. \sqrt{4+400}

Add. \sqrt{404}

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6 0
3 years ago
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Which of the following is the most appropriate unit to describe the rate at which a person reads?
alukav5142 [94]
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</span>
6 0
3 years ago
Need help finding the area of the shaded region. Round to the nearest tenth. Need help ASAP
Sergio [31]

Answer:   294.4 m²

<u>Step-by-step explanation:</u>

Separate the shaded region into two parts:

  1. The section containing the central angle of 230° (360° - 130°)
  2. The triangle with sides 11.1, 11.1 & 20.12 (use Law of Cosines)

1.\ Area(A)=\pi\ r^2\ \bigg(\dfrac{\theta}{360}\bigg)\\\\\\.\qquad \qquad =\pi(11.1)^2\bigg(\dfrac{230}{360}\bigg)\\\\\\.\qquad \qquad =247.3

2.\ \text{Use Law of cosines to find the length of the third side.}\\\text{ Then use Heron's formula to find the Area of the triangle.}\\\\s=\dfrac{11.1+11.1+20.12}{2}=21.16\\\\\\A=\sqrt{s(s-a)(s-b)(s-c)}\\\\.\ =\sqrt{21.16(21.16-11.1)(21.16-11.1)(21.16-20.12)}\\\\.\ =\sqrt{2227}\\\\.\ =47.1

Area of shaded region = Area of (1) + Area of (2)

                                      =    247.3     +      47.1

                                      =               294.4

8 0
4 years ago
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