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Roman55 [17]
3 years ago
14

8x + 6 < 54 Please help

Mathematics
2 answers:
Alexandra [31]3 years ago
8 0
X<6
download an app called photomath
katen-ka-za [31]3 years ago
7 0

Answer:

x < 6

Step-by-step explanation:

Isolate the variable (x). Do the opposite of PEMDAS. What you do to one side, you do to the other.

First, subtract 6 from both sides.

8x + 6 (-6) < 54 (-6)

8x < 48

Next, divide 8 from both sides

(8x)/8 < (48)/8

x < 48/8

x < 6

x < 6 is your answer

~

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Answer:

(x + 1)^{2}  + (y + 11)^{2} = 34

Step-by-step explanation:

First find the midpoint of the segment  from (2, -6) to (-4, -16)

((2 - 4)/2, (-6 - 16)/2)

(-2/2, -22/2)

(-1, -11) is the midpoint and is the center of the circle

Now find the distance between the 2 given points and divide by 2 to get the radius of the circle

d = \sqrt{(2 + 4)^{2} + (-6 + 16)^{2}  }

  = \sqrt{6^{2} + 10^{2}  }

  = \sqrt{36 + 100}

  = \sqrt{136}

r = \sqrt{136}/2

r^{2} = 136/4 = 34

Equation of circle with center at (-1, -11) and radius = \sqrt{34} is

(x + 1)^{2}  + (y + 11)^{2} = 34

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3 years ago
Determine whether a triangle with the given vertices is a right triangle.
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2 years ago
Pls help me hurry I need help!!
Citrus2011 [14]

C. 9 square roots of 2

To answer this question, it's super important that you understand the ratio of sides for special triangles. This triangle in particular, a 45-45-90 triangle, has a ratio between the legs and hypotenuse of 1:1:

Since we are given the value of the hypotenuse, we know that the value of the two sides multiplied by \sqrt{2} will be 18. Knowing this, we can write out an equation:

u*\sqrt{2} = 18

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<u>Multiply both sides by </u>\sqrt{2}<u> in order to get rid of the root in the denominator:</u>

u = \frac{18*\sqrt{2}}{\sqrt{2} * \sqrt{2}}

u = \frac{18*\sqrt{2} }{2}

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2 years ago
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<span>More please help me solve the problem The stars have different brightness. The brightest stars are Grade 1 and the least degree luimnoase stars 6. The brightness of the stars shrinks 2.5 times with the passing from one grade to another. Whenever the stars are brighter Grade 1, Grade 6 than the stars?</span>
8 0
2 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
3 years ago
Read 2 more answers
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